Alternating and interpolated MZVs:

Let us recall, again, the notions of alternating MZVs, and of multiple zeta star values and multiple zeta-half values, which will be useful imminently. Given a tuple $ (k_1,k_2,\ldots ,k_d) $ of positive integers, and a tuple $ (\varepsilon _1,\varepsilon _2,\ldots ,\varepsilon _d) \in \{ \pm 1 \}^d $ , with $ (k_d,\varepsilon _d) \neq (1,1) $ , we define the alternating MZV (or Euler sum) with signs $ \varepsilon _1,\ldots ,\varepsilon _d $ as follows,

One then streamlines the notation by suppressing the $ \varepsilon _i $ ’s and writing $ \overline {k_i} $ if $ \varepsilon _i = -1 $ , and just $ k_i $ if $ \varepsilon _i = 1 $ otherwise. For example

An alternating MZV can be written as an iterated integral in the following way

(A.1) $$ \begin{align} \zeta\bigg(\begin{matrix} \varepsilon_1,\varepsilon_2,\ldots,\varepsilon_d \\ k_1,k_2,\ldots,k_d \end{matrix}\bigg) = (-1)^d I^{\mathfrak{m}}(0; \eta_1, \{0\}^{k_1-1}, \eta_2, \{0\}^{k_2-1}, \ldots, \eta_d, \{0\}^{k_d-1} ;1) , \end{align} $$

where $ \eta _i = \varepsilon _i \varepsilon _{i+1} \cdots \varepsilon _d $ .

Next, we have the interpolated multiple zeta values $ \zeta ^r(k_1,\ldots ,k_d) $ introduced by Yamamoto [Reference Yamamoto39],

For example, $ \zeta ^r(a,b,c) = \zeta (a,b,c) + r \zeta (a+b,c) + r \zeta (a, b+c) + r^2 \zeta (a+b+c) $ . In the case $ r = 0 $ , only the term with all $ \circ _i = \text{","} $ survives, and so $ \zeta ^0(k_1,\ldots ,k_d) = \zeta (k_1,\ldots ,k_d) $ . When $ r = 1 $ , then we have $ \zeta ^1(k_1,\ldots ,k_d) = \zeta ^\star (k_1,\ldots ,k_d) $ , where the multiple zeta star value (MZSV) is originally defined as

and arises by replacing the strict inequalities between $ n_i, n_{i+1} $ with a nonstrict inequality. For $ r = {{1 }/{2}} $ , we then obtain a new variant ‘mid-way’ between $ \zeta $ and $ \zeta ^\star $ , called the multiple zeta-half value.

This formalism can be extended to allow for alternating interpolated MZVs, by replacing $ {`+}\text{'} $ , above with $ {`}{\oplus} \text{'} $ , where $ a \oplus b $ denotes addition of the absolute values and multiplication of the bars viewed as signs (i.e. if $ k \in \mathbb {Z}_{>0} $ , we have $ \left |k\right | = \left |\smash {\overline {k}}\right | = k $ with $ \operatorname {\mathrm {sgn}}(\overline {k}) = -1 $ and $ \operatorname {\mathrm {sgn}}(k) = 1 $ ). In particular, for $ \alpha ,\beta \in \mathbb {Z}_{>0} $ , we have $ \alpha \oplus \beta = \alpha + \beta $ , $ \alpha \oplus \overline {\beta } = \overline {\alpha } \oplus \beta = \overline {\alpha + \beta } $ and $ \overline {\alpha } \oplus \overline {\beta } = \alpha + \beta $ . Then, for example, we have the following interpolated alternating MZV

$$ \begin{align*} \zeta^r(a, \overline{b},\overline{c}) & {} = \zeta(a,\overline{b},\overline{c}) + r \zeta(a\oplus \overline{b},\overline{c}) + r \zeta(a, \overline{b}\oplus\overline{c}) + r^2 \zeta(\overline{a}\oplus \overline{b} \oplus \overline{c}) \\ & {} = \zeta(a,\overline{b},\overline{c}) + r \zeta(\overline{a+b},\overline{c}) + r \zeta(a, b+c) + r^2 \zeta(\overline{a+b+c}) . \end{align*} $$

The case $ r = {{1 }/{2}} $ of alternating interpolated MZVs is a convenient way of formulating Zhao’s [Reference Zhao41] generalised 2-1 Theorem, as we will see below.

A.1 Stuffle antipode

We define $ G^\star $ , the generating series of $ \zeta ^\star (\{2\}^a, 4 ,\{2\}^b) $ , $ G $ , a related generating series for $ \zeta (\{2\}^a, 4, \{2\}^b) $ , and $ S^\star $ the generating series of $ \zeta ^\star (\{2\}^n) $ as follows.

Then from [Reference Li31, Equation 2.4] (in the special case, $ a_2 = z_4 $ , $ a_1 = a_3 = z_2 $ ), we have that

(A.2) $$ \begin{align} G^\star(x,y) = G(y,x) \ast S^\star(x) \ast S^\star(y) . \end{align} $$

This is an identity in the stuffle algebra; in particular, it automatically lifts to a motivic identity since the stuffle product is known to be motivic (see [Reference Racinet34] or [Reference Soudères37]). Moreover, it is well-known (or readily verifiable by factoring the generating series as a product, see, for example [Reference Borwein, Bradley and Broadhurst2, Equation 36], and [Reference Zagier40, Equation 44]) that

$$\begin{align*}\sum_{n=0}^\infty \zeta^\star(\{2\}^n) x^{2n} = \frac{\pi x}{\sin(\pi x)} , \text{ and } \sum_{n=0}^\infty \zeta(\{2\}^n) x^{2n} = \frac{\sin(\mathrm{i} \pi x)}{\mathrm{i} \pi x} . \end{align*}$$

By solving Equation (A.2) for $ G(y,x) $ , and extracting the coefficient of $ x^{2a} y^{2b} $ , we obtain the following explicit formula for $ \zeta (\{2\}^a, 4, \{2\}^b) $ in terms of similar $ \zeta ^\star $ values,

(A.3) $$ \begin{align} \zeta(\{2\}^a,4,\{2\}^b) = \sum_{n=0}^a \sum_{m=0}^b (-1)^{m+n} \zeta^\star(\{2\}^m, 4, \{2\}^n) \zeta(\{2\}^{a-n}) \zeta(\{2\}^{b-m}) . \end{align} $$

A similar identity holds for $ 4 $ replaced by any value $ k $ ; these identities give the precise version of the stuffle antipode result

$$\begin{align*}\zeta^{\mathfrak{l}}(k_1,\ldots,k_d) = (-1)^{d+1} \zeta^{\star,\mathfrak{l}}(k_d,\ldots,k_1) \end{align*}$$

considered in [Reference Glanois18, Lemma 4.2.2]. Moreover, since the stuffle algebra identity and the evaluations of $ \zeta (\{2\}^n) $ , $ \zeta ^\star (\{2\}^n) $ are motivic ([Reference Brown5, Lemma 3.4], [Reference Glanois18, Lemma 4.4.3]), Equation (A.3) lifts automatically to a motivic version.

A.2 Generalised 2-1 Theorem

We now recall the generalised 2-1 Theorem, established by Zhao [Reference Zhao41], which evaluates each $ \zeta ^\star $ value in terms of a certain alternating $ \zeta ^{{1 }/{2}} $ value. It is more convenient – and indeed has a closer connection with the goal – to write the generalised 2-1 Theorem in the block decomposition form given in [Reference Charlton12, Lemma 3.1].

Let $ \mathbf {\underline {s}} = (s_1,\ldots ,s_k) $ be a sequence of MZV arguments, and let $ \mathbf {\underline {B}} = (\ell _1,\ldots ,\ell _n) $ be the corresponding block decomposition (see Section 2.3). Write

$$\begin{align*}\widetilde{\,x\,} = \begin{cases} x & \text{if}\ x\ \text{odd}, \\ \overline{x} & \text{if}\ x\ \text{even} \end{cases} \end{align*}$$

Recall: $ \overline {x} $ denotes that the argument $ x $ in an alternating MZV has sign $ -1 $ . Then

$$\begin{align*}\zeta^\star(\mathbf{\underline{s}}) = \varepsilon(\mathbf{\underline{s}}) \cdot 2^n \zeta^{{1 }/{2}}(\widetilde{\ell_1 - 2}, \widetilde{\ell_2}, \ldots, \widetilde{\ell_n}) , \end{align*}$$

where $ \varepsilon (\mathbf {\underline {s}}) = 1 $ if $ s_1 = 1 $ and $ \varepsilon (\mathbf {\underline {s}}) = -1 $ if $ s_1 \geq 2 $ , and if $ \ell _1-2 = 0 $ , one should neglect this argument. This follows by combining Zhao’s generalised 2-1 Theorem [Reference Zhao41], which involves a certain recursively constructed sequence of indices $ \mathbf {\underline {s}}^{(i)} $ , with the description of the final such index string $ \mathbf {\underline {s}}^{(k)} $ , given in [Reference Charlton12, Lemma 3.1], in terms of the block decomposition. This final string supplies the $ \zeta ^{{1 }/{2}} $ arguments in Zhao’s formulation of the 2-1 Theorem.

In our case, we want to apply this to $ \zeta ^\star (\{2\}^a, 4, \{2\}^b) $ . The block decomposition of $ (\{2\}^a, 4, \{2\}^b) $ is given by $ (2a+3,1,2b+2) $ , and we therefore have

$$\begin{align*}\zeta^\star(\{2\}^a, 4, \{2\}^b) = -2^3 \zeta^{{1 }/{2}}(2a + 1, 1, \overline{2b+2}) . \end{align*}$$

Then expanding out, by definition of the interpolated $ \zeta ^{{1 }/{2}} $ , we have

(A.4) $$ \begin{align} \nonumber \zeta^\star(\{2\}^a, 4, \{2\}^b) = & -2 \zeta(\overline{2a + 2b + 4}) - 4 \zeta(2a+1, \overline{2b + 3}) \\ & - 4 \zeta(2a + 2, \overline{2b + 2}) - 8 \zeta(2a + 1, 1, \overline{2b+2}) . \end{align} $$

This reduces our task of evaluating $ \zeta ^\star (\{2\}^a, 4, \{2\}^b) $ to understanding certain explicit depth 3 alternating MZVs.

A.3 The parity theorem in depth 3

The parity theorem for MZVs states roughly that an MZV of weight $ w $ and depth $ d $ can be reduced to a combination of lower depth MZVs and products, when $ w \not \equiv d \quad\pmod {2} $ . In particular, an MZV of depth 3 and even weight is reducible. This claim actually also holds for alternating MZVs, via the parity theorem for multiple polylogarithms [Reference Panzer33], as $ -1 $ is its own multiplicative inverse.

An explicit version of the depth 3 parity theorem is given for the multiple polylogarithm functions $ \operatorname {\mathrm {Li}}_{n_1,n_2,n_3}(z_1,z_2,z_3) $ in [Reference Panzer33, Equation 4.3]. By specialising to $ z_i = \pm 1 $ , we recover the claimed reduction of depth 3 alternating MZVs, for any choice of signs $ z_i $ (encoded with a ‘bar’ over the corresponding argument $ \overline {n_i} $ , if $ z_i = -1 $ ), as follows. Namely, if $ \alpha + \beta + \gamma $ is even and $ \gamma \neq 1 $ (although $ \gamma = \bar {1} $ is okay), then with $\zeta (0) = \zeta (\overline {0}) = -\frac {1}{2}$ by convention, and stuffle-regularisation if necessary (see [Reference Ihara, Kaneko and Zagier25] for the notion of regularisation, and Remark A.1 below for the behaviour in this case) when $\zeta (1)$ appears, we have

$$ \begin{align*} \zeta(\alpha,\beta,\gamma) &= \frac{1}{2} \zeta(\alpha) \big( \zeta(\beta,\gamma) - (-1)^{\left|\beta\right|+\left|\gamma\right|} \zeta(\beta,\gamma) \big) - \zeta(\beta,\alpha) \zeta(\gamma) \delta_{\left|\gamma\right| \text{even}} \\ &\quad - \frac{1}{2} \zeta(\alpha \oplus \beta, \gamma) + \frac{1}{2} \zeta(\beta \oplus \gamma, \alpha) \\ &\quad + \sum_{\substack{2s + \nu + \mu = \left|\smash{\alpha}\right| \\ s, \mu, \nu \geq 0}} (-1)^{\left|\beta\right| + \left|\gamma\right| + \mu+ \nu} \zeta(\operatorname{\mathrm{sgn}}(\alpha\beta\gamma) \diamond 2s)\binom{-\left|\beta\right|}{\mu} \binom{-\left|\gamma\right|}{\nu} \zeta(\beta\oplus \mu, \gamma \oplus \nu) \\ &\quad + \sum_{\substack{2s + \nu + \mu = \left|\smash{\beta}\right| \\ s, \mu, \nu \geq 0}} (-1)^{\gamma + \mu} \zeta(\operatorname{\mathrm{sgn}}(\alpha\beta\gamma) \diamond 2s)\binom{-\left|\gamma\right|}{\mu} \binom{-\left|\alpha\right|}{\nu} \zeta(\gamma \oplus \mu) \zeta( \alpha \oplus \nu) \\ &\quad + \sum_{\substack{2s + \nu + \mu = \left|\smash{\gamma}\right| \\ s, \mu,\nu \geq 0}} \zeta(\operatorname{\mathrm{sgn}}(\alpha\beta\gamma) \diamond 2s) \binom{-\left|\beta\right|}{\mu} \binom{-\left|\alpha\right|}{\nu} \zeta(\beta \oplus \mu, \alpha \oplus \nu) \, \end{align*} $$

To avoid abuse of notation, we define , and , to give the corresponding decoration for signed arguments.

Now specialise to $ \alpha = 2a + 1, \beta = 1, \gamma = \overline {2b+2} $ . We can simplify various binomial coefficients and powers of $ -1 $ , using $ \binom {-k}{\ell } = (-1)^\ell \binom {\ell + k-1}{k-1} $ , and expand out the second summation into its two nontrivial terms $ (s,\mu ,\nu ) = (0, 1, 0), (0, 0, 1) $ . After doing so, and inserting the result into Equation (A.4), we note some simplifications. Firstly, the term $ -4\zeta (2a+2,\overline {2b+2}) $ in the $ \zeta ^{{1 }/{2}} $ cancels with one from the depth 3 reduction; secondly, the term $ -4\zeta (2a+1,\overline {2b+3}) $ combines with one from the depth 3 reduction to produce

$$ \begin{align*} & -4\zeta(2a+1,\overline{2b+3}) - 4\zeta(\overline{2b+3},2a+1) \\ &\quad = -4\zeta(2a+1)\zeta(\overline{2b+3}) + 4 \zeta(\overline{2a+2b+4}). \end{align*} $$

Overall, this produces the following evaluation for $ \zeta (\{2\}^a, 4, \{2\}^b) $ , as the first main stepping stone, with stuffle-regularisation applied where necessary

(A.5) $$ \begin{align} \nonumber \zeta^\star(\{2\}^a, 4, \{2\}^b) &= 2 \zeta(\overline{2a + 2b + 4}) + 8 \zeta(1, 2a+1) \zeta(\overline{2b+2}) - 8 \zeta(2a + 1)\zeta(1, \overline{2b + 2}) \\ \nonumber &\quad + 4 (2b + 1) \zeta(\overline{2b +3}) \zeta(2a + 1) - 4 (2a + 1) \zeta(\overline{2b + 2}) \zeta(2a + 2) \\ &\quad + 8 \sum_{\substack{2s + \nu + \mu = 2a+1 \\ s, \mu, \nu \geq 0}} \zeta(\overline{2s}) \binom{\nu+(2b+1)}{\nu} \zeta(1 + \mu, \overline{2b + 2 + \nu}) \\ \nonumber &\quad - 8 \sum_{\substack{2s + \nu + \mu = 2b+2 \\ s, \mu, \nu \geq 0}} \zeta(\overline{2s}) \binom{\nu + (2a)}{\nu} \zeta(1 + \mu, 2a + 1 + \nu) . \end{align} $$

Remark A.1 (Independence of regularisation).

Let us note here that the shuffle-regularised and stuffle-regularised versions of this formula agree (and, indeed, also for the depth 3 reduction, with $ c \neq 1 $ ), and are independent of the regularisation parameter; we may therefore switch to the shuffle-regularisation at $ T = 0 $ for later convenience. This is expected since we are reducing a convergent triple zeta value.

More precisely, this is because terms with a single trailing 1 are equal under either regularisation, and in the case $ a = 0 $ , the single term $ 8\zeta (\overline {2b+1})\zeta (1,1) $ with two trailing 1’s, which arises, arising from $ (s,\nu ,\mu ) = (b+1, 0, 0) $ in the last sum, cancels with the corresponding term on the second line. Otherwise, when $ a = 1 $ , the regularisation parameter $ T $ in terms arising from $ \nu = 0 $ in the last sum (with $ \zeta ^{\ast ,T} $ explicitly denoting the stuffle-regularised version with $ \zeta ^{\ast ,T} (1) = T $ ),

$$\begin{align*}\sum_{\substack{2s + \mu = 2b + 2 \\ s \geq 0,\mu> 0}} \zeta(\overline{2s}) \zeta^{\ast,T}(1 + \mu, 1) = \sum_{\substack{2s + \mu = 2b + 2 \\ s \geq 0,\mu > 0}} \zeta(\overline{2s}) \big( T \zeta(1 + \mu) - \zeta(1, 1+\mu) - \zeta(2+\mu) \big) \end{align*}$$

can be seen to cancel with that arising from the terms

$$ \begin{align*} & 8 \zeta^{\ast,T}(1) \zeta(1,\overline{2b+2}) + 4 (2b+1) \zeta(\overline{2b+3}) \zeta^{\ast,T}(1) \\ &\quad = 8 T \zeta(1, \overline{2b+2}) + 4 (2b+1) \zeta(\overline{2b+3}) T. \end{align*} $$

In fact, this cancellation is equivalent to the following reduction which follows from the depth-parity theorem in depth 2 (see [Reference Panzer33, Equation 3.5]):

(A.6) $$ \begin{align} \zeta(1, \overline{2b+2}) = -\sum_{\substack{2s+k = 2b + 3 \\ s \geq 0, k \geq 2}} \zeta(\overline{2s}) \zeta(k) + \frac{2b+1}{2} \zeta(\overline{2b+3}) . \end{align} $$

A.4 Shuffle-regularisation and dihedral symmetries

Now let us take advantage of the shuffle-regularisation in earnest. Define $ \zeta _\ell (k_1,\ldots ,k_d) $ by inserting $ \ell $ leading 0’s at the start of the iterated integral representation of $ \zeta (k_1,\ldots ,k_d) $ given in Equation (A.1) (see also Equations (2.1) and (B.1)), and write for the shuffle regularisation thereof (see Remark 2.2 above), with . Then we have the regularisation formula (see, for example [Reference Brown10, Section 5.1 R2], and the obvious generalisation to alternating MZVs in [Reference Glanois18, Equation 2.28])

In particular, using this, we can write

Substituting these and Equation (A.6) into Equation (A.5) gives us the following

Let us finally note that the term $ 8\zeta (1,2a+1) \zeta (\overline {2b+2}) $ cancels with the $ s = b+1 $ term of the last sum. In particular, we obtain the second stepping stone in our quest to evaluate $ \zeta ^{\star }(\{2\}^a, 4, \{2\}^b) $ . Namely

(A.7)

Henceforth, we shall always use the shuffle regularisation with , and will therefore drop the extraneous from our notation. This regularisation is consistent with the regularisation normally used in the motivic framework (c.f. Remark 2.2). Moreover, we will check in Section B.2 that this reduction is indeed motivic.

Now, we recall from [Reference Glanois18, Corollary 4.2.6] that depth $ p $ alternating zeta star values satisfy a dihedral symmetry of order $ p + 1 $ , modulo products and lower depth. More precisely therein, this symmetry is phrased in terms of so-called multiple zeta star-star values, which incorporate the lower depth terms, making the symmetry hold already modulo products. In particular, in our case, we claim that

$$ \begin{align*} \zeta_{2k-1}(1,\overline{2\ell}) &\equiv \zeta(\overline{2\ell},\overline{2k}) \quad\pmod{\text{products}} \\ \zeta_{2k}(1,2\ell+1) &\equiv \zeta(2\ell+1,2k+1) \quad\pmod{\text{products}} \, \end{align*} $$

The depth 1 terms in these cases are reducible, as the weight is even. We will not actually use the implicit form of the dihedral symmetry established by Glanois which immediately produces the above; instead, guided by the Glanois’s proof, we will establish an exact version in this depth 2 case. However, let us point out some technical issue which apparently occurs when considering the octagon relation, in an attempt to derive a so-called hybrid relation (Theorem 4.2.3 in [Reference Glanois18]), a key part of the proof of the dihedral symmetry.

Remark A.2 (Regularisation in the octagon relation).

The octagon relation for level $ N = 2 $ multiple zeta values (i.e. alternating MZVs) is obtained by integrating a word in $ e_0, e_1, e_{-1} $ around the following closed loop.

However, one must take into account the tangential base points, and how they are transformed under the Möbius transformation which cyclically maps the segments $ (0,1) \mapsto (-1,0) \mapsto (\infty ,-1) \mapsto (1,\infty ) $ . More precisely, the Möbius transformation

$$\begin{align*}f(z) = \frac{z-1}{z+1} \end{align*}$$

transforms the segments as indicated, and therefore the integral $ \int _{(-1,0)} $ is related to the integral $ \int _{(0,1)} $ via a suitable pullback. However, note that the straight line path

$$ \begin{align*} \mathrm{dch} \colon [0,1] &\to [0,1] \\ t &\mapsto t \end{align*} $$

with tangential base points $ \overrightarrow {0}_{1} $ and $ \overrightarrow {1}_{-1} $ is transformed into the path

$$ \begin{align*} (f^\ast \mathrm{dch})(t) \colon [0,1] &\to [-1,0] \\ t &\mapsto \frac{t-1}{t+1} \end{align*} $$

with tangential base points $ \overrightarrow {-1}_{2} $ and $ \overrightarrow {0}_{-\frac {1}{2}} $ . In particular, the semicircular integrals evaluate in the following way

$$\begin{align*}\int_{\overrightarrow{0}_{-\frac{1}{2}}}^{\overrightarrow{0}_1} \frac{\mathrm{d}t}{t} = \log(2) - \mathrm{i} \pi . \end{align*}$$

So the octagon relation actually takes the form

$$ \begin{align*} e^{\big(-\frac{\mathbb{L}^{\mathfrak{m}}}{2} + \log^{\mathfrak{m}}(2)\big) e_{-1}} & \Phi^{\mathfrak{m}}(e_0, e_{-1}, e_1) e^{\big(-\frac{\mathbb{L}^{\mathfrak{m}}}{2} + \log^{\mathfrak{m}}(2)\big) e_{0}} \Phi^{\mathfrak{m}}(e_0, e_{1}, e_{-1}) \\ & \cdot e^{\big(-\frac{\mathbb{L}^{\mathfrak{m}}}{2} + \log^{\mathfrak{m}}(2)\big) e_{1}} \Phi^{\mathfrak{m}}(e_\infty, e_{1}, e_{-1}) e^{\big(-\frac{\mathbb{L}^{\mathfrak{m}}}{2} + \log^{\mathfrak{m}}(2)\big) e_{\infty}} \Phi^{\mathfrak{m}}(e_\infty, e_{-1}, e_1) = 1 , \end{align*} $$

where $ e_\infty $ is defined, such that $ e_0 + e_1 + e_\infty + e_{-1} = 0 $ .

This change should not render Glanois’s hybrid identity invalid, as the derivation of the hybrid identity mainly requires the octagon relation modulo products, and these additional terms largely cancel out.

We now turn to the derivation of the exact identities which verify our earlier claim that

$$ \begin{align*} \zeta_{2k-1}(1,\overline{2\ell}) &\equiv \zeta(\overline{2\ell},\overline{2k}) \quad\pmod{\text{products}} \\ \zeta_{2k}(1,2\ell+1) &\equiv \zeta(2\ell+1,2k+1) \quad\pmod{\text{products}} . \end{align*} $$

We treat the first, as the second is exactly analogous; we will, nevertheless, give the full identity in each case. Firstly, apply shuffle regularisation to

$$\begin{align*}\zeta_{z-1}(\overline{\alpha}, \beta) + \zeta_{z-1}(\beta, \overline{\alpha}) \, \end{align*}$$

to obtain

$$\begin{align*}= (-1)^{z-1} \sum_{i+j = z-1} \binom{i+\alpha-1}{i} \binom{j+\beta-1}{j} \big\{ \zeta(\overline{i+\alpha}, j+\beta) + \zeta(j+\beta, \overline{i+\alpha}) \big\} \, \end{align*}$$

Note, we have combined the two original sums by switching $ i \leftrightarrow j $ in the second sum. By the stuffle product (switching to stuffle regularisation is okay, as there is at most a single trailing 1), we have

$$\begin{align*}= (-1)^{z-1} \sum_{i+j = z-1} \binom{i+\alpha-1}{i} \binom{j+\beta-1}{j} \big\{ \zeta(j+\beta)\zeta(\overline{i+\alpha}) - \zeta(\overline{\alpha+\beta+z-1}) \big\} . \end{align*}$$

On the other hand, apply the shuffle antipode [Reference Goncharov21, Equation 29]

$$ \begin{align*} & (-1)^N I(a; x_N, \ldots, x_1; b) + I(a; x_1,\ldots,x_N; b) \\ &\quad + \sum_{i=1}^{N-1} (-1)^{N-i} I(a; x_1, \ldots, x_i; b) I(a; x_{N}, \ldots, x_{i+1}; b) = 0 \end{align*} $$

which effectively reverses the differential forms in an iterated integral $ I(a, x_1,\ldots ,x_N; b) $ , modulo explicit products terms, to

$$\begin{align*}\zeta_{z-1}(\overline{\alpha}, \beta) = I(0; \{0\}^{z-1}, -1, \{0\}^{\alpha-1}, 1, \{0\}^{\beta-1}; 1), \end{align*}$$

and we find

$$ \begin{align*} & \zeta_{z-1}(\overline{\alpha}, \beta) + (-1)^{z-1+\alpha+\beta} \zeta_{\beta-1}(\overline{\alpha}, \overline{z}) \\ &\quad =I(0; \{0\}^{z-1}, -1, \{0\}^{\alpha-1}, 1, \{0\}^{\beta-1}; 1) \\ & \qquad + (-1)^{z-1+\alpha+\beta} I(0; \{0\}^{\beta-1}, 1, \{0\}^{\alpha-1}, -1, \{0\}^{z-1}; 1) \\ &\quad = -\sum_{i=0}^{\alpha-1} (-1)^{z-1 + \alpha + i} \binom{z+i-1}{i} \binom{\alpha+\beta-2-i}{\beta-1} \zeta(\overline{z+i}) \zeta(\alpha+\beta-1-i) . \end{align*} $$

Finally, take the difference of these two identities, and set $ \alpha = 2\ell , \beta = 1, z = 2k $ , we then obtain the dihedral symmetry we claimed

$$ \begin{align*} \zeta_{2k-1}(1, \overline{2\ell}) - \zeta(\overline{2\ell}, \overline{2k}) &= -\sum_{i=0}^{2k-1} \binom{i+2\ell-1}{i} \big\{ \zeta(\overline{i+2\ell}) \zeta(2k-i) - \zeta(\overline{2k+2\ell}) \big\} \\ &\quad - \sum_{i=0}^{2\ell-1} (-1)^i \binom{2k+i-1}{i} \zeta(\overline{2k+i}) \zeta(2\ell-i) .\end{align*} $$

A slightly more concise version of this is obtained by extending the sums to negative indices – where the binomial coefficients vanish – in order to combine them into one, and explicitly summing the coefficient of $ \zeta (\overline {2k+2\ell }) $ . This puts the identity in a form closer to that which one could directly check/derive with the motivic derivations, namely

(A.8) $$ \begin{align} \nonumber \zeta_{2k-1}(1, \overline{2\ell}) - \zeta(\overline{2\ell}, \overline{2k}) & = \binom{2k+2\ell-1}{2k-1} \zeta(\overline{2k+2\ell}) \\ &\quad - \sum_{\substack{r = 1}}^{2k+2\ell-2} \bigg( (-1)^r \binom{r-1}{2k-1} + \binom{r-1}{2\ell-1} \bigg) \zeta(\overline{r}) \zeta(2k+2\ell-r) . \end{align} $$

In an analogous way, we find the explicit form of the dihedral symmetry in the other case to be

(A.9) $$ \begin{align} \nonumber \zeta_{2k}(1, 2\ell+1) - {} & \zeta(2\ell+1,2k+1) = {} - \zeta(2) \delta_{k=\ell=0} - \binom{2k+2\ell+1}{2\ell+1} \zeta(2k+2\ell+2) \\ & + \sum_{\substack{r = 1}}^{2k+2\ell} \bigg( (-1)^r \binom{r-1}{2k} + \binom{r-1}{2\ell} \bigg) \zeta(r) \zeta(2k+2\ell+2-r) \, \end{align} $$

Here, the term $ \delta _{k=\ell =0} $ accounts for the difference in shuffle- and stuffle-regularisation in the case $ \zeta (1,1) $ .

Both of these identities are easily verified to be motivic, either by direct calculation via $ \operatorname {\mathrm {D}}_{2r+1} $ , or by noting that the ingredients – namely, the shuffle and stuffle products, and the regularisation – are themselves motivic in nature.

A.5 Generalised doubling identity

The final ingredient we require for our evaluation is one of the so-called generalised doubling identities, as described in [Reference Blümlein, Broadhurst and Vermaseren1, Section 4], and [Reference Zhao42, Section 14.2.5] (be aware these references use the opposite MZV convention).

In depth 2, the relevant relation is already given explicitly by Zhao, and states (with either shuffle or stuffle regularisation) that

$$ \begin{align*} & \zeta(s,t) + \zeta(\overline{s},\overline{t}) \\ &\quad = \sum_{i=1}^s \binom{s+t-i-1}{t-1} 2^{1+i-s-t} \zeta(i, s+t-i) + \sum_{i=1}^t \binom{s+t-i-1}{s-1} 2^{1+i-s-t} \zeta(s+t-i,i) \\ &\qquad - \sum_{i=1}^t \binom{s+t-i-1}{s-1} \big\{ \zeta(s+t-i,i) + \zeta(\overline{s+t-i},i) \big\} - \binom{s+t-1}{s} 2^{1-s-t} \zeta(s+t) .\end{align*} $$

We then flip $ \zeta (\overline {a},b) $ to $ \zeta (b,\overline {a}) $ using the stuffle product, rewrite the double zeta sums that lack powers of 2 using the shuffle regularisation as before, and simplify the resulting coefficient of $ \zeta (s+t) $ (The power of 2 does indeed just disappear!) This gives the equivalent identity

$$ \begin{align*} & \zeta(\overline{s},\overline{t}) + (-1)^t \zeta_{t-1}(1,\overline{s}) \\ & \quad = \sum_{i=1}^s \binom{s+t-i-1}{t-1} 2^{1+i-s-t} \zeta(i, s+t-i) + \sum_{i=1}^t \binom{s+t-i-1}{s-1} 2^{1+i-s-t} \zeta(s+t-i, i) \\ &\qquad - \zeta(s,t) + (-1)^t \zeta_{t-1}(s,1) - \sum_{i=1}^t \binom{s+t-i-1}{s-1} \zeta(\overline{s+t-i})\zeta(i) - \binom{s+t-1}{s} \zeta(s+t) . \end{align*} $$

Finally, we note that upon substituting $ t = 2k, s = 2\ell $ , we can solve this identity simultaneously with Equation (A.8) to obtain expressions for both $ \zeta (\overline {2\ell },\overline {2k}) $ and $ \zeta _{2k-1}(1,\overline {2\ell }) $ individually as combinations of classical depth 2 MZVs and products. In particular, we have established the following proposition (after substituting an expression for $ (-1)^t \zeta _{t-1}(s,1) = \zeta _{2k-1}(2\ell ,1) \equiv \zeta (2k,2\ell ) \quad\pmod {\text {products}} $ using the shuffle antipode, or via a further dihedral symmetry, and simplifying).

Proposition A.3 (Galois descent of $ \zeta (\overline {2\ell },\overline {2k}) $ ).

The alternating double zeta value $ \zeta (\overline {2\ell },\overline {2k}) $ enjoys a Galois descent to classical depth 2 MZVs as follows

(A.10) $$ \begin{align} \nonumber \zeta(\overline{2\ell},\overline{2k}) = & \sum_{i=2}^{2k + 2\ell - 2} 2^{-i} \bigg\{ \binom{i-1}{2k-1} \zeta(2k+2\ell-i,i) + \binom{i-1}{2\ell-1} \zeta(i,2k+2\ell-i) \bigg\} \\ & -\zeta(2\ell,2k) + \sum_{r=2}^{2k+2\ell-2} (-2)^{-r} \binom{r-1}{2k-1} \zeta(r) \zeta(2k+2\ell-r) \\ \nonumber & - 2^{-2k-2\ell} \bigg\{ 2 \binom{2k+2\ell-2}{2k-1} + \binom{2k+2\ell-1}{2k-1} \bigg\} \zeta(2k + 2\ell) . \end{align} $$

Moreover, by a direct calculation, since $ \operatorname {\mathrm {D}}_{2r+1} $ is a tensor product of single-zeta values in this case, we see Proposition A.3 (and the generalised doubling identity itself) lifts to the motivic level. This is checked in detail in Section B.1.

Remark A.4. It is clear from the generalised 2-1 Theorem that $ \zeta (\overline {2\ell },\overline {2k}) $ descends to a combination of classical MZVs; we can, in fact, easily give an explicit formula

$$\begin{align*}\zeta(\overline{2\ell},\overline{2k}) = \frac{1}{4} \zeta^\star(1,\{2\}^{\ell-1}, 3, \{2\}^{k-1}) - \frac{1}{2} \zeta(2k+2\ell) . \end{align*}$$

However, it is certainly not clear from this expression that depth 2 classical MZVs suffice, and so this would not help us in evaluating $ \zeta (\{2\}^a, 4, \{2\}^b) $ in any useful manner. However, we do obtain an evaluation for $ \zeta ^\star (1,\{2\}^{\ell -1}, 3, \{2\}^{k-1}) $ by substituting Proposition A.3 into the above.

Moreover, since $ \zeta (1,1,4,6) $ is – according to the Data Mine [Reference Blümlein, Broadhurst and Vermaseren1] – a combination of depth 2 alternating MZVs and products

$$ \begin{align*} \zeta (1,1,4,6) = & \frac{64}{9} \zeta (\overline{3},\overline{9}) +\frac{371}{144} \zeta (3,9) +3 \zeta (2) \zeta (3,7) +\frac{3}{2} \zeta (4) \zeta (3,5) -\frac{3131}{144} \zeta (9) \zeta (3) \\ &+\frac{107}{24} \zeta (5) \zeta (7) +10 \zeta (2) \zeta (7) \zeta (3) +\frac{7}{2} \zeta (2) \zeta (5)^2 -\frac{1}{2} \zeta (4) \zeta (5) \zeta (3) \\ &-\frac{9}{4} \zeta (6) \zeta (3)^2 +\frac{\zeta (3)^4}{12} -\frac{117713 }{132672}\zeta (12) , \end{align*} $$

but apparently irreducible as a classical MZV, one cannot, in general, expect the Galois descent to always respect the depth. However, as pointed out in Remark 5.2, one has – assuming the homological version of the Broadhurst-Kreimer Conjecture [Reference Broadhurst and Kreimer3] (see also Conjecture 1.1 above) – that the depth of an alternating MZV after Galois descent should be at most twice the original; here, the Galois descent of $ \zeta (\overline {3},\overline {9}) $ involving classical MZVs up to depth 4 corroborates this.

By substituting Proposition A.3 into Equation (A.8), and this result into Equation (A.7), we establish that $ \zeta ^\star (\{2\}^a, 4, \{2\}^b) $ and (via Equation (A.3)) that $ \zeta (\{2\}^a, 4, \{2\}^b) $ are both expressible in terms of only classical double zeta values.

A.6 Generating series

In order to extract an explicit useable identity for $ \zeta (\{2\}^a, 4, \{2\}^b) $ , we will convert everything to a generating series identity as a route to simplifying the resulting combinations.

Let us introduce the following generating series, whose names originate from Zagier’s evaluation of $ \zeta (\{2\}^a, 3, \{2\}^b) $ [Reference Zagier40], and some related generating series of even zeta values. The generating series of odd MZVs, and alternating odd MZVs, are given by

$$ \begin{align*} A(z) &= \sum_{r=1}^\infty \zeta(2r+1) z^{2r} = \psi(1) - \frac{1}{2} \psi(1 - z) - \frac{1}{2} \psi(1+z), \\ B(z) &= \sum_{r=1}^\infty (-\zeta(\overline{2r+1})) z^{2r} = \sum_{r=1}^\infty (1 - 2^{-2r}) \zeta(2r+1) z^{2r} = A(z) - A(\tfrac{z}{2}) , \end{align*} $$

where $ \psi (x) = \frac {\mathrm {d}}{\mathrm {d}x} \log \Gamma (x) $ is the digamma function, the logarithmic derivative of the Gamma function. We keep with the choice established by Zagier taking negatives in the series for alternating MZVs. The generating series for even versions, using the convention that $ \zeta (0) = \zeta (\overline {0}) = -\frac {1}{2} $ , are given by

$$ \begin{align*} E(z) &= \sum_{r=0}^\infty \zeta(2r) z^{2r-1} = -\frac{\pi}{2} \cot(\pi z) , &\quad F(z) &= \sum_{r=0}^\infty (-\zeta(\overline{2r})) z^{2r-1} = \frac{\pi}{2} \csc(\pi z) . \\ \widetilde{E}(z) &= \sum_{r=1}^\infty \zeta(2r) z^{2r-1} = \frac{1}{2x}-\frac{\pi}{2} \cot(\pi z) , &\quad \widetilde{F}(z) &= \sum_{r=0}^\infty (-\zeta(\overline{2r})) z^{2r-1} = -\frac{1}{2x} + \frac{\pi}{2} \csc(\pi z) . \end{align*} $$

The $ \widetilde {\bullet } $ versions, which are missing the polar term, will be convenient later. Let us introduce the following double zeta generating series

$$ \begin{align*} D_{\mathrm{ev}}(x,y) &= \sum_{a,b=1}^\infty \zeta(2a, 2b) x^{2a-1} y^{2b-1} , \quad D_{\mathrm{od}}(x,y) = \sum_{a,b=1}^\infty \zeta(2a + 1, 2b + 1) x^{2a} y^{2b} , \\ D(x,y) &= { \sum_{\substack{a,b=2 \\ a \equiv b \quad\pmod{2}}}^\infty \zeta(a,b) x^{a-1} y^{b-1} = D_{\mathrm{od}}(x,y) + D_{\mathrm{ev}}(x,y). } \end{align*} $$

As an intermediate step, let us also introduce the following generating series to capture the shuffle-regularised zetas appearing in Equation (A.7), and in the dihedral symmetries in Equations (A.8) and (A.9), as well as the alternating zeta values as part of the Galois descent result

$$ \begin{align*} K_{\mathrm{alt}}(x,y) &= \sum_{a,b=1}^\infty \zeta(\overline{2a},\overline{2b}) x^{2a-1} y^{2b-1} , & \quad & K_{\mathrm{ev}}(x,y) = \sum_{a,b=1}^\infty \zeta_{2b-1}(1,\overline{2a}) x^{2a-1} y^{2b-1} , \\ K_{\mathrm{od}}(x,y) &= \sum_{a,b=0}^\infty \zeta_{2b}(1,2a + 1) x^{2a} y^{2b}. \end{align*} $$

Note that we sum from $ a,b = 0 $ in $ K_{\mathrm {od}} $ but will restrict this to start from $ a,b = 1 $ in $ D_{\mathrm {od}} $ , on account of the well-known reductions of $ \zeta (1, 2b+1) $ , and $ \zeta (2a+1, 1) $ .

Generating series for Equation (A.7).

To obtain the generating series $ G^\star (x,y) $ , we sum the left-hand side of Equation (A.7) weighted by $ x^{2a} y^{2b} $ over all $ a, b \geq 0 $ .

We find then that the generating series of the first term on the right-hand side is

$$ \begin{align*} \sum_{a,b=0}^\infty \zeta(\overline{2a+2b+4}) x^{2a} y^{2b} &= \sum_{r=0}^\infty \sum_{s=0}^r \zeta(\overline{2r+4}) x^{2r-2s} y^{2s} \\ &= \sum_{r=0}^\infty \zeta(\overline{2r+4}) \sum_{s=0}^r x^{2r-2s} y^{2s} \\ &= \sum_{r=0}^\infty \zeta(\overline{2r+4}) \cdot \frac{x^{2r+2} - y^{2r+2}}{x^2 - y^2} \\&= - \frac{y \widetilde{F}(x) - x \widetilde{F}(y)}{x y(x^2 - y^2)} . \end{align*} $$

Likewise, the second leads to

$$ \begin{align*} \sum_{a,b=0}^\infty (2a+1) \zeta(2a+2) \zeta(\overline{2b+2}) x^{2a} y^{2b} &= \sum_{a=0}^\infty (2a+1) \zeta(2a+2) x^{2a} \cdot \sum_{b=0}^\infty \zeta(\overline{2b+2}) y^{2b} \\ &= \frac{1}{y} \frac{\mathrm{d} \widetilde{E}(x) }{\mathrm{d}x} \cdot \widetilde{F}(y). \end{align*} $$

The third and fourth terms are readily summed to give

$$ \begin{align*} & \sum_{a,b=0}^\infty \zeta(2a+1) \cdot \sum_{k=1}^{b+1} \zeta(2k+1) \zeta(\overline{2b+2-2k}) x^{2a} y^{2b} = -\frac{1}{y} A(x) A(y) F(y) \\ & \sum_{a,b=0}^\infty \sum_{s=0}^a \zeta(\overline{2s}) \zeta_{2a+1-2s}(1, \overline{2b+2}) x^{2a} y^{2b} = - \frac{1}{y} F(x) K_{\mathrm{ev}}(y,x). \end{align*} $$

The last term requires a little care, as the summand $ \zeta _{0}(1,2a+1) $ does not appear, and so must be discounted via $ K_{\mathrm {od}}(x,0) $ , the constant-in- $ y $ term of $ K_{\mathrm {od}}(x,y) $ . That is

$$ \begin{align*} & \sum_{a,b=0}^\infty \sum_{s=0}^b \zeta(\overline{2s}) \zeta_{2b+2-2s}(1, 2a+1) x^{2a} y^{2b} = -\frac{1}{y} F(y) ( K_{\mathrm{od}}(x,y) - K_{\mathrm{od}}(x,0) ) . \end{align*} $$

This gives us

$$ \begin{align*} G^\star(x,y) = & \frac{8}{y} A(x) A(y) F(y) + \frac{8}{y} F(x) K_{\mathrm{ev}}(y,x) + \frac{8}{y} F(y) (K_{\mathrm{od}}(x,y) - K_{\mathrm{od}}(x,0)) \\ & - 2 \cdot \frac{y \widetilde{F}(x) - x \widetilde{F}(y)}{x y(x^2 - y^2)} + \frac{4}{y} \frac{\mathrm{d} \widetilde{E}(x) }{\mathrm{d}x} \cdot \widetilde{F}(y). \end{align*} $$

Generating series for dihedral identities:

The generating series of the form given by summing $ \sum _{k,\ell =1}^\infty (\bullet ) y^{2k-1} x^{2\ell -1} $ (note the order of the variables), over the left-hand side of Equation (A.8) is simply just $ K_{\mathrm {ev}}(x,y) - K_{\mathrm {alt}}(x,y) $ . Then

$$ \begin{align*} \sum_{k,\ell=1}^\infty \binom{2k+2\ell-1}{2k-1} \zeta(\overline{2k+2\ell}) y^{2k-1} x^{2\ell-1} &= \sum_{r=2}^\infty \zeta(\overline{2r}) \sum_{k=1}^{r-1} \binom{2r-1}{2k-1} y^{2k-1} x^{2r-2k-1} \\ &= \sum_{r=2}^\infty \zeta(\overline{2r}) \Big\{ {-}y^{2r} + \frac{y}{2} (y-x)^{2r-1} + \frac{y}{2} (y+x)^{2r-1} \Big\} \\ &= -\frac{1}{2x} \Big( \widetilde{F}(y-x) + \widetilde{F}(y+x) - 2 \widetilde{F}(y) \Big). \end{align*} $$

Similarly

$$ \begin{align*} & \sum_{k,\ell=1}^\infty \sum_{r=1}^{2k+2\ell-2} (-1)^r \binom{r-1}{2k-1} \zeta(\overline{r})\zeta(2k+2\ell-r) \cdot y^{2k-1} x^{2\ell-1} \\ &\quad = \sum_{k,\ell=1}^\infty \bigg( {-}\sum_{r=0}^{k+\ell-2} \binom{2r}{2k-1} \zeta(\overline{2r+1}) \zeta(2k+2l-2r-1) \\ &\qquad + \sum_{r=1}^{k+\ell-1} \binom{2r+1}{2k-1} \zeta(\overline{2r}) \zeta(2k+2\ell-2r) \bigg) y^{2k-1} x^{2\ell-1} \\ & \quad = \sum_{r=0}^\infty \sum_{s=1}^\infty \sum_{k=1}^r \bigg( {-} \binom{2r}{2k-1} \zeta(\overline{2r+1}) \zeta(2s-1) + \binom{2r-1}{2k-1} \zeta(\overline{2r}) \zeta(2s) \bigg)y^{2k-1} x^{2s+2r-2k-1}. \end{align*} $$

The sum over $ k $ can be evaluated explicitly, and (taking care with the $ r = 0 $ terms) one obtains

$$ \begin{align*} = \sum_{r,s=1}^\infty & \frac{1}{2} \big( (y-x)^{2r} - (y+x)^{2r} \big) \zeta(\overline{2r+1}) \zeta(2s-1) x^{2s-2} \\[-1ex] & + \sum_{r,s=1}^\infty \frac{1}{2} \big( (y-x)^{2r-1} + (y+x)^{2r-1} \big) \zeta(\overline{2r}) \zeta(2s) x^{2s-1} \\ = -\frac{1}{2} & \big( B(y-x) - B(y+x) \big) A(x) - \frac{1}{2} \big( \widetilde{F}(y-x) + \widetilde{F}(y+x) \big) \widetilde{E}(x) . \end{align*} $$

Likewise, one finds

$$ \begin{align*} & \sum_{k,\ell=1}^\infty \sum_{r=1}^{2k+2\ell-2} \binom{r-1}{2\ell-1} \zeta(\overline{r})\zeta(2k+2\ell-r) \cdot y^{2k-1} x^{2\ell-1} \\ &\quad = \frac{1}{2} \big( B(y-x) - B(y+x) \big) A(y) + \frac{1}{2} \big( \widetilde{F}(y-x) - \widetilde{F}(y+x) \big) \widetilde{E}(y) , \end{align*} $$

which essentially amounts to switching $ x \leftrightarrow y $ , and switching the sign between the two terms.

Overall, one obtains

$$ \begin{align*} & K_{\mathrm{ev}}(x,y) - K_{\mathrm{alt}}(x,y)\\ &\quad = \frac{1}{2} (A(x) - A(y))(B(x-y) - B(x+y)) + \frac{1}{2} \widetilde{E}(y) (\widetilde{F}(x-y) + \widetilde{F}(x+y)) \\ &\qquad - \frac{1}{2} \widetilde{E}(x) (\widetilde{F}(x-y) - \widetilde{F}(x+y)) - \frac{\widetilde{F}(x+y) - \widetilde{F}(x-y) - 2 \widetilde{F}(y)}{2x}. \end{align*} $$

In exactly the same way, one finds for the second dihedral identity Equation (A.9) – taking care with the terms $ \zeta (1,2b+1) $ and $ \zeta (2a+1,1) $ missing from $ D_{\mathrm {od}}(x,y) $ – that

$$ \begin{align*} & K_{\mathrm{od}}(x,y) - D_{\mathrm{od}}(x,y)\\ &\quad = 2\zeta(2) + \frac{1}{2} (A(x) - A(y))(A(x) - A(x-y) + A(y) - A(x+y)) \\ &\qquad + \frac{1}{2} ( \widetilde{E}(x) - \widetilde{E}(y) ) (\widetilde{E}(x) + \widetilde{E}(y) - \widetilde{E}(x-y)) + \frac{1}{2} \big( \widetilde{E}(x) + \widetilde{E}(y) \big) \widetilde{E}(x+y) \\ &\qquad - \frac{\widetilde{E}(x-y) + \widetilde{E}(x+y)}{2x} - \frac{\widetilde{E}(y)}{y}. \end{align*} $$

One may observe from this, that

$$\begin{align*}K(x,0) = \frac{1}{4} \zeta(2) - \frac{3}{8x^2} - \frac{1}{2} A(x)^2 + \Big( \frac{\pi^2}{4} + E(x)^2 \Big) \Big( \frac{1}{2} + \frac{\sin(2\pi x)}{2 \pi x} \Big) . \end{align*}$$

Generating series for Equation (A.10):

Now, we compute the generating series for the identity from Proposition A.3. Again, taking $ \sum _{k,\ell =1}^\infty (\bullet )y^{2k-1} x^{2\ell -1} $ , note the variable order, we find the left-hand side to be just $ K_{\mathrm {alt}}(x,y) $ . The binomial times double zeta terms can be summed as follows

$$ \begin{align*} & \sum_{k,\ell=1}^\infty \sum_{i=2}^{2k+2\ell-2} 2^{-i} \binom{i-1}{2k-1} \zeta(2k+2\ell-i,i) \cdot y^{2k-1} x^{2\ell-1} \\ &\quad = \sum_{r,s=1}^\infty \Big\{ \zeta(2r,2s) \cdot 2^{-2s} \sum_{k=1}^s \binom{2s-1}{2k-1} x^{2r + 2s - 2k-1} y^{2k-1} \\ &\qquad + \zeta(2r + 1, 2s + 1) \cdot 2^{-2s-1} \sum_{k=1}^s \binom{2s}{2k-1} x^{2r + 2s + 1 - 2k} y^{2k-1} \Big\} \\ &\quad = \frac{1}{4} \sum_{r,s=1}^\infty \Big\{ \zeta(2r,2s) x^{2r-1} \Big\{ \Big( \frac{x+y}{2} \Big)^{2s-1} - \Big( \frac{x-y}{2} \Big)^{2s-1} \Big\} \\ &\qquad + \zeta(2r+1,2s+1) x^{2r} \Big\{ \Big( \frac{x+y}{2} \Big)^{2s} - \Big( \frac{x-y}{2} \Big)^{2s} \Big\} \Big\} \\ &\quad = \frac{1}{4} \Big( D\Big(x, \frac{x+y}{2}\Big) - D\Big(x, \frac{x-y}{2}\Big) \Big), \end{align*} $$

The other terms may be handled similarly. Overall, we find the generating series identity

$$ \begin{align*} K_{\mathrm{alt}}(x,y) = & \frac{1}{4} \Big( D\Big(x, \frac{x+y}{2}\Big) - D\Big(x, \frac{x-y}{2}\Big) + D\Big(\frac{x+y}{2},y\Big) - D\Big(-\frac{x-y}{2},y\Big) \Big) - D_{\mathrm{ev}}(x,y) \\ & + \frac{1}{4} \Big( \widetilde{E}\Big(\frac{x+y}{2}\Big) - \widetilde{E}\Big(\frac{x-y}{2}\Big) \Big) \widetilde{E}(x) - \frac{1}{4} A(x) \Big( A\Big( \frac{x+y}{2} \Big) - A\Big( \frac{x-y}{2} \Big) \Big) \\ & + \frac{3x - y}{4x(x-y)} \widetilde{E}\Big(\frac{x-y}{2}\Big) - \frac{3x + y}{4x(x+y)} \widetilde{E}\Big( \frac{x+y}{2} \Big) + \frac{1}{2x} \widetilde{E}\Big(\frac{y}{2} \Big). \end{align*} $$

A.7 Explicit evaluations for $ \zeta ^\star (\{2\}^a, 4, \{2\}^b) $

After substituting the previous generating series into the expression for $ G^\star (x,y) $ given via Equation (A.7), performing a nontrivial amount of trigonometric manipulation, we find

$$ \begin{align*} G^\star(x,y) &= \frac{2 F(x)}{y} \Big( {-}D\Big(\frac{x-y}{2}, x\Big) - D\Big(y, -\frac{x-y}{2}\Big) + D\Big(y, \frac{x+y}{2}\Big) + D\Big(\frac{x+y}{2}, x\Big) \Big) \\ &\quad - \frac{8 F(x)}{y} D_{\mathrm{ev}}(y,x) + \frac{8 F(y)}{y} D_{\mathrm{od}}(x,y) \\ &\quad + \frac{2 F(x)}{y} \Big\{ - A(y) \cdot \Big( A\Big( \frac{x+y}{2} \Big) - A\Big( \frac{x-y}{2} \Big) \Big) + 2 \big( A(x) - A(y) \big) \cdot \big( B(x+y) - B(x-y) \big) \Big\} \\ &\quad + \frac{4 F(y)}{y} \Big\{ A(x) \cdot \big(A(x) - 2 A(y) \big) + \big(A(x) - A(y) \big) \cdot \big( A(x) + A(y) - A(x-y) - A(x+y) \big) \Big\} \\ &\quad - 3 \zeta(2) \cdot \frac{1 - y \widetilde{E}(y)}{y^2} \cdot \sec\Big(\frac{\pi(x-y)}{2} \Big) \sec\Big( \frac{\pi(x+y)}{2} \Big) - \frac{2 \big( x F(x) - y F(y) \big)}{y^2(x^2-y^2)} \\ &\quad + \frac{4 F(x)}{y} \bigg\{ \widetilde{E}(x) \Big( \frac{1}{y} + \widetilde{E}(x+y) - \widetilde{F}(x-y) \Big) - \frac{\widetilde{E}(\tfrac{1}{2} (x+y))}{x+y} + \frac{\widetilde{E}(\tfrac{1}{2} (x-y))}{x-y} \bigg\} \\ &\quad + \frac{4 F(y)}{y} \bigg\{ 2\zeta(2) + \widetilde{E}(y)^2 - \frac{2 \widetilde{E}(y)}{y} + \frac{x \widetilde{E}(x) - y \widetilde{E}(y)}{x^2 - y^2} \\ & \quad - \frac{(x+y) \widetilde{E}(x-y) - (x-y) \widetilde{E}(x+y)}{2 x y} \bigg\}. \end{align*} $$

If one so desires, the following explicit formula for the individual coefficient $\zeta ^\star (\{2\}^a, 4, \{2\}^b)$ can be extracted from the above.

Theorem A.6 (Evaluation of $\zeta ^\star (\{2\}^a, 4, \{2\}^b)$ via double zeta valuesFootnote ² ).

Write as shorthand $ \zeta (\overline {n}) = -(1-2^{1-n})\zeta (n) $ , and employ the conventions that $ \zeta (0) = \zeta (\overline {0}) = -\frac {1}{2} $ and $ \zeta (1) = 0 $ . However, no further regularisation is necessary. Denote by $ E_n $ the $ n $ -th Euler number, given as the coefficients of $ \operatorname {sech}(t) = \sum _{n=0}^\infty \frac {E_n}{n!} t^n . $ Then for any $ a, b \in \mathbb {Z}_{\geq 0} $ , the following evaluation holds, where we assume all summation variables start from $ 0 $ ,

A.8 Explicit evaluations for $ \zeta (\{2\}^a, 4, \{2\}^b) $

By substituting the expression for $ G^\star (x,y) $ into Equation (A.2), and finding $ G(x,y) $ via

$$\begin{align*}G(x,y) = \sum_{a,b=0}^\infty (-1)^{a+b} \zeta(\{2\}^a, 4, \{2\}^b) x^{2a} y^{2b} = G^\star(y,x) \frac{\sin(\pi x)}{\pi x} \frac{\sin(\pi y)}{\pi y} , \end{align*}$$

we obtain the following explicit expression for the generating series,

$$ \begin{align*} G(x,y) &= \frac{\sin(\pi x)}{\pi x^2 y} \Big( {-}D\Big(-\frac{x-y}{2}, x\Big) - D\Big(x, \frac{x-y}{2}\Big) + D\Big(x, \frac{x+y}{2}\Big) + D\Big(\frac{x+y}{2}, y\Big) \Big) \\ &\quad - \frac{4\sin(\pi x)}{\pi x^2 y} D_{\mathrm{ev}}(x,y) + \frac{\sin(\pi y)}{\pi x^2 y} D_{\mathrm{od}}(y,x) \\ &\quad - \frac{\sin(\pi x)}{\pi x^2 y} \Big\{ A(x) \cdot \Big( A\Big( \frac{x+y}{2} \Big) - A\Big( \frac{x-y}{2} \Big) \Big) + 2 \big( A(x) - A(y) \big) \cdot \big( B(x+y) - B(x-y) \big) \Big\} \\ &\quad - \frac{\sin(\pi y)}{\pi x^2 y} \Big\{ A(y) \cdot \big(2A(x) - A(y) \big) + \big(A(x) - A(y) \big) \cdot \big( A(x) + A(y) - A(x-y) - A(x+y) \big) \Big\} \\ &\quad - 3 \zeta(2) \cdot \frac{1 - x \widetilde{E}(x)}{x^2} \cdot \sec\Big(\frac{\pi(x-y)}{2} \Big) \sec\Big( \frac{\pi(x+y)}{2} \Big) \frac{\sin(\pi x)}{\pi x}\frac{\sin(\pi y)}{\pi y} \\ & \quad - \frac{1}{x^2(x^2-y^2)} \Big( \frac{\sin(\pi x)}{\pi x} - \frac{\sin(\pi y)}{\pi y} \Big)\\ &\quad + \frac{2\sin(\pi x)}{\pi x^2 y} \bigg\{ \widetilde{E}(y) \Big( \frac{1}{x} + \widetilde{E}(x+y) + \widetilde{F}(x-y) \Big) - \frac{\widetilde{E}(\tfrac{1}{2} (x+y))}{x+y} + \frac{\widetilde{E}(\tfrac{1}{2} (x-y))}{x-y} \bigg\} \\ &\quad + \frac{2\sin(\pi y)}{\pi x^2 y} \bigg\{ 2\zeta(2) + \widetilde{E}(x)^2 - \frac{2 \widetilde{E}(x)}{x} + \frac{x \widetilde{E}(x) - y \widetilde{E}(y)}{x^2 - y^2} \\ & \quad + \frac{(x+y) \widetilde{E}(x-y) - (x-y) \widetilde{E}(x+y)}{2 x y} \bigg\} . \end{align*} $$

If one desires, the following explicit formula for the individual coefficient $ \zeta (\{2\}^a, 4, \{2\}^b) $ can be extracted from the above.

Theorem A.7 (Evaluation of $\zeta (\{2\}^a, 4, \{2\}^b)$ via double zeta valuesFootnote ³ ).

Write as shorthand $ \zeta (\overline {n}) = -(1-2^{1-n})\zeta (n) $ , and employ the conventions that $ \zeta (0) = \zeta (\overline {0}) = -\frac {1}{2} $ and $ \zeta (1) = 0 $ . However, no further regularisation is necessary. Denote by $ E_n $ the $ n $ -th Euler number, given as the coefficients of $ \operatorname {sech}(t) = \sum _{n=0}^\infty \frac {E_n}{n!} t^n . $ Then for any $ a, b \in \mathbb {Z}_{\geq 0} $ , the following evaluation holds, where we assume all summation variables start from $ 0 $ ,

In particular, we obtain the following corollary on the reduction of $ \zeta (\{2\}^a, 4, \{2\}^b) $ modulo products. In essence, it extracts those double zeta terms above, which are not multiplied by a power of $ \pi $ .

Corollary A.8. Modulo decomposables (i.e. products of MZVs), the following evaluation holds

Framework of alternating motivic MZVs:

We note, here, that the motivic MZV framework of Section 2.1 generalises readily to the case of alternating motivic MZVs. For the technical details of this, we refer to [Reference Glanois18, Reference Glanois19]; the most important points are the comodule of alternating motivic MZVs $ \mathcal {H}^{(2)} $ is obtained by extending Definition 2.1 to allow $ a_i \in \{0, \pm 1 \} $ (although functoriality in a useful form only applies when all $ a_i \in \{ 0, 1 \}$ ). Then for a tuple $ (k_1,\ldots ,k_d) $ of positive integers, and $ (\varepsilon _1,\ldots ,\varepsilon _d) \in \{\pm 1\}^d $ of signs, and $ \ell \geq 0 $ , we define the motivic alternating MZV by

(B.1)

where $ \eta _i = \varepsilon _i \varepsilon _{i+1} \cdots \varepsilon _d $ , mimicking the integral representation of alternating MZVs in Equation (A.1). One can again streamline the notation by dropping the $ \varepsilon _i $ ’s and writing $ \overline {k_i} $ if $ \varepsilon _i = -1 $ , and just $ k_i $ if $ \varepsilon _i = 1 $ . Then and $ \mathcal {L}^{(2)} = \mathcal {A}^{(2)}_{>0} / \mathcal {A}^{(2)}_{>0}\mathcal {A}^{(2)}_{>0} $ define the obvious extensions of the Hopf algebra and the Lie coalgebra of irreducibles. The coaction $ \Delta \colon \mathcal {H}^{(2)} \to \mathcal {A}^{(2)} \otimes \mathcal {H}^{(2)} $ is defined by the same formula as in Equation (2.2), and the infinitesimal derivations $ \operatorname {\mathrm {D}}_r \colon \mathcal {H}^{(2)} \to \mathcal {L}^{(2)}_r \otimes \mathcal {H}^{(2)} $ , with $ \mathcal {L}^{(2)}_r $ the weight $ r $ component of $ \mathcal {L}^{(2)} $ , are given by the same formula as in Equation (2.3).

For alternating motivic MZVs, $ \operatorname {\mathrm {D}}_1 $ plays a nontrivial role, as the weight 1 alternating MZV $ \zeta ^{\mathfrak {m}}(\overline {1}) = \log ^{\mathfrak {m}}(2) $ is nonzero. The analogue of Brown’s [Reference Brown5] characterisation of $ \ker \operatorname {\mathrm {D}}_{<N} $ in the alternating case is given by Glanois as follows.

Theorem B.1 (Glanois, Corollary 2.4.5 [Reference Glanois18]).

Let $ N \geq 1 $ , and denote by $ \operatorname {\mathrm {D}}_{<N} = \bigoplus _{1 \leq 2r+1 < N} \operatorname {\mathrm {D}}_{2r+1} $ . Then in weight $ N $ , the kernel of $ D_{<N} $ on alternating motivic MZVs is one dimensional:

$$\begin{align*}\ker \operatorname{\mathrm{D}}_{<N} \cap \mathcal{H}^{(2)}_N = \begin{cases} \mathbb{Q} \zeta^{\mathfrak{m}}(\overline{1}) = \mathbb{Q} \log^{\mathfrak{m}}(2) & \text{if}\ N = 1 \\ \mathbb{Q} \zeta^{\mathfrak{m}}(N) & \text{if}\ N>1 . \end{cases} \end{align*}$$

Since the identities we wish to lift involve alternating MZV terms in a nontrivial way, we necessarily have to use Glanois’s criterion to verify the motivic lift, even if, as it happens, $ \operatorname {\mathrm {D}}_1 = 0 $ in each case.

B.1 Motivic version of Proposition A.3

We prove the following proposition which claims that Proposition A.3 lifts to a motivic version.

Proposition B.2 (Motivic Galois descent of $ \zeta ^{\mathfrak {m}}(\overline {2\ell },\overline {2k}) $ ).

The alternating motivic double zeta value $ \zeta ^{\mathfrak {m}}(\overline {2\ell },\overline {2k}) $ enjoys a Galois descent to classical depth 2 motivic MZVs as follows

(B.2) $$ \begin{align} \nonumber \zeta^{\mathfrak{m}}(\overline{2\ell},\overline{2k}) = & \sum_{i=2}^{2k + 2\ell - 2} 2^{-i} \bigg\{ \binom{i-1}{2k-1} \zeta^{\mathfrak{m}}(2k+2\ell-i,i) + \binom{i-1}{2\ell-1} \zeta^{\mathfrak{m}}(i,2k+2\ell-i) \bigg\} \\ & -\zeta^{\mathfrak{m}}(2\ell,2k) + \sum_{r=2}^{2k+2\ell-2} (-2)^{-r} \binom{r-1}{2k-1} \zeta^{\mathfrak{m}}(r) \zeta^{\mathfrak{m}}(2k+2\ell-r) \\ \nonumber & - 2^{-2k-2\ell} \bigg\{ 2 \binom{2k+2\ell-2}{2k-1} + \binom{2k+2\ell-1}{2k-1} \bigg\} \zeta^{\mathfrak{m}}(2k + 2\ell) . \end{align} $$

Computation of $ {\operatorname {\mathrm {D}}}_{2r+1} \zeta ^{\mathfrak {m}}(\overline {2a},\overline {2b})$ and ${\operatorname {\mathrm {D}}}_{2r+1} \zeta ^{\mathfrak {m}}(2a, 2b)$

We see that only the following subsequences can contribute to the motivic coaction. This is because any subsequence must start or end one of the three nonzero entries; one then checks whether the length $ 2r+1 $ subsequences which start/end at these points actually contribute

We find

$$ \begin{align*} & \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}(\overline{2a}, \overline{2b}) \\ &\quad = -\delta_{a \leq r} \zeta_{2r + 1 - 2a}^{\mathfrak{l}}(\overline{2a}) \otimes \zeta(2a + 2b - 2r - 1) + \delta_{b\leq r} \zeta_{2r + 1 - 2b}^{\mathfrak{l}}(\overline{2b}) \otimes \zeta^{\mathfrak{m}}(2a + 2b - 2r - 1) \\ &\quad = \bigg( \binom{2r}{2a-1} - \binom{2r}{2b-1} \bigg) \zeta^{\mathfrak{l}}(\overline{2r+1}) \otimes \zeta(2a + 2b - 2r - 1) \end{align*} $$

The binomial factors should a prior retain the delta factors, but they can be removed as the binomials vanish already for the complementary condition. The corresponding result holds for $ \zeta ^{\mathfrak {m}}(2a, 2b) $ by removing all bars from the above result

$$ \begin{align*} &\operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}({2a}, {2b}) = \bigg( \binom{2r}{2a-1} - \binom{2r}{2b-1} \bigg) \zeta^{\mathfrak{l}}({2r+1}) \otimes \zeta(2a + 2b - 2r - 1). \end{align*} $$

Computation of $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}(2a+1, 2b+1) $

We see that only the following subsequences can contribute to the motivic coaction. This is because any subsequence must involve one of the three nonzero entries; one then checks whether the length $ 2r+1 $ subsequences which start/end at these points actually contribute

We find

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}({2a+1}, {2b+1}) & = \delta_{a=r} \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta^{\mathfrak{m}}(2a + 2b + 1 - 2r) \\ &\quad + \Big( {-} \delta_{a\leq r} \zeta_{2r-2a}^{\mathfrak{l}}(2a+1) + \delta_{b \leq r} \zeta_{2r-2b}^{\mathfrak{l}}(2b+1) \Big) \otimes \zeta^{\mathfrak{m}}(2a+2b+1-2r) \\ & = \bigg( \delta_{a=r} - \binom{2r}{2a} + \binom{2r}{2b} \bigg) \zeta^{\mathfrak{l}}({2r+1}) \otimes \zeta(2a + 2b + 1 - 2r) . \end{align*} $$

Computation of $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}(p, q) $ , $ p + q $ even

We note that the two cases above can be combined to give the following, for $ p + q $ even

$$\begin{align*}\operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}(p,q) = \bigg( \delta_{2r+1 = p} + (-1)^p \binom{2r}{p-1} - (-1)^q \binom{2r}{q-1} \bigg) \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta(p+q - 2r - 1) . \end{align*}$$

Verification of Proposition B.2

The claim that $ \operatorname {\mathrm {D}}_{2r+1} $ of both sides agree is equivalent to the following putative identity among binomial coefficients, when $ 1 \leq r \leq k+\ell -1 $ , which arises after projecting $ \zeta ^{\mathfrak {l}}(2r+1) \otimes \zeta ^{\mathfrak {m}}(2k+2\ell -2r-1) \mapsto 1 $ .

$$ \begin{align*} 0 &\overset{?}{=} (1-2^{-2r}) \bigg( \binom{2r}{2\ell-1} - \binom{2r}{2k-1} \bigg) \\ &\quad + \sum_{i=2}^{2k+2\ell-2} 2^{-i} \binom{i-1}{2k-1} \bigg( \delta_{2k+2\ell-i = 2r+1} + (-1)^i \binom{2r}{2k+2\ell-i-1} - (-1)^{i} \binom{2r}{i-1} \bigg) \\ &\quad + \sum_{i=2}^{2k+2\ell-2} 2^{-i} \binom{i-1}{2\ell-1} \bigg( \delta_{i = 2r+1} + (-1)^i \binom{2r}{i-1} - (-1)^{i} \binom{2r}{2k+2\ell-i-1} \bigg) \\ &\quad - \bigg( \binom{2r}{2l-1} - \binom{2r}{2k-1} \bigg) \\ &\quad + (-2)^{-(2r+1)} \binom{2r}{2k-1} + (-2)^{-(2k+2\ell-2r-1)} \binom{2k+2\ell-2r-2}{2k-1}. \end{align*} $$

After some simplification of the right-hand side, and reindexing the sums, we find that the claim is equivalent to the following

$$ \begin{align*} 0 \overset{?}{=} & -2^{-2r-1} \bigg( \binom{2r}{2\ell-1} - \binom{2r}{2k-1} \bigg) \\ & + (-2)^{-2k} \sum_{i=0}^{2\ell-2} (-2)^{-i} \binom{i+2k-1}{2k-1} \bigg( \binom{2r}{2\ell-i-1} - \binom{2r}{2k+i-1} \bigg) \\ & + (-2)^{-2\ell} \sum_{i=0}^{2k-2} (-2)^{-i} \binom{i+2k-1}{2\ell-1} \bigg( \binom{2r}{2\ell+i-1} - \binom{2r}{2k-i-1} \bigg). \end{align*} $$

This is verified to be exactly 0 from Lemma 4.2 of Section 4. With that, we have finished the proof of Proposition B.2.

B.2 Motivic version of Equation (A.7)

We prove the following proposition, which claims that Equation (A.7) lifts to a motivic version.

Proposition B.3. The following identity holds among motivic multiple zeta (star) values

(B.3)

Computation of $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}_{2a+1}(1, \overline {2b+2}) $

We see that only the following subsequences can contribute to the motivic coaction. This is because any subsequence must start or end at one of the three nonzero entries; one then checks whether the length $ 2r+1 $ subsequences which start/end at these points actually contribute

Hence, we have

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}_{2a+1}(1,\overline{2b+2}) & = \delta_{r\leq a} \zeta^{\mathfrak{l}}_{2r}(1) \otimes \zeta^{\mathfrak{m}}_{2a+1-2r}(\overline{2b+2}) - \delta_{r \leq b} \zeta^{\mathfrak{l}}_{2r}(1) \otimes \zeta^{\mathfrak{m}}_{2a+1}(\overline{2b+2-2r}) \\ & = \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( {-} \delta_{r\leq a} \binom{2a + 2b + 2 - 2r}{2b + 1} + \delta_{r \leq b} \binom{2a + 2b + 2 - 2r}{2a} \bigg) \\ &\quad \cdot \zeta^{\mathfrak{m}}(\overline{2a + 2b+ 3 - 2r}) . \hspace{-3em} \end{align*} $$

And, in particular, $ \operatorname {\mathrm {D}}_1 = 0 $ .

Computation of $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}_{2b+2}(1, 2a+1) $

Similarly, only the following subsequences can contribute to the motivic coaction

Hence, we have

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}_{2b+2}(1,2a+1) & = \delta_{r\leq b+1} \zeta^{\mathfrak{l}}_{2r}(1) \otimes \zeta^{\mathfrak{m}}_{2b+2-2r}(2a+1) - \delta_{r \leq a-1} \zeta^{\mathfrak{l}}_{2r}(1) \otimes \zeta^{\mathfrak{m}}_{2b+2}(2a+1-2r) \\ & = \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( {-} \delta_{r\leq b+1} \binom{2a+2b+2-2r}{2a + 1} + \delta_{r \leq a-1} \binom{2a+2b+2 - 2r}{2b + 1} \bigg) \\ & \quad \cdot \zeta^{\mathfrak{m}}(2a + 2b+ 3 - 2r) .\hspace{-3em} \end{align*} $$

Computation of $ \operatorname {\mathrm {D}}_{2r+1} $ of right-hand side of Equation (B.3)

With the above two computations of the motivic coaction on the double zeta values in Equation (B.3), we can readily compute the rest of the coaction using the derivation property of $ \operatorname {\mathrm {D}}_{2r+1} $ , namely, $ \operatorname {\mathrm {D}}_{2r+1} X Y = (1 \otimes Y) \operatorname {\mathrm {D}}_{2r+1}X + (1 \otimes X) \operatorname {\mathrm {D}}_{2r+1} Y $ , as well as the fact that $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}(N) = \delta _{N=2r+1} \zeta ^{\mathfrak {l}}(N) \otimes 1 $ . Note also the first two terms on the right-hand side of Equation (B.3) are products of even zetas, and so do not contribute. So we find

$$ \begin{align*} & \operatorname{\mathrm{D}}_{2r+1} (\text{RHS Equation}~({\textrm{B.3}}))\\ &\quad = 8 \delta_{r=a} \zeta^{\mathfrak{l}}(2r+1) \otimes \sum_{k=1}^{b+1} \zeta^{\mathfrak{m}}(2k+1) \zeta^{\mathfrak{m}}(\overline{2b+2-2k}) \\ &\qquad + 8 \delta_{r \leq b+1} \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta^{\mathfrak{m}}(2a+1) \zeta^{\mathfrak{m}}(\overline{2b + 2 - 2r}) \\ &\qquad - 8 \sum_{s = 0}^{a} \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( {-} \delta_{r\leq a-s} \binom{2a + 2b + 2 - 2r - 2s}{2b+1} + \delta_{r \leq b} \binom{2a + 2b + 2 - 2r - 2s}{2b + 1 - 2r} \bigg) \\ &\qquad \cdot \zeta^{\mathfrak{m}}(\overline{2a - 2s + 2b + 3 - 2r}) \zeta^{\mathfrak{m}}(\overline{2s}) \\ &\qquad - 8 \sum_{s=0}^b \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( \delta_{r\leq b-s+1} \binom{2a + 2b + 2 - 2r - 2s}{2a} - \delta_{r \leq a-1} \binom{2a + 2b + 2 - 2r - 2s}{2a - 2r} \bigg) \\ &\qquad \cdot \zeta^{\mathfrak{m}}(2a + 2b - 2s + 3 - 2r) \zeta^{\mathfrak{m}}(\overline{2s}). \end{align*} $$

Computation of $ \operatorname {\mathrm {D}}_{2r+1} $ of left-hand side of Equation (B.3)

We compute the derivation $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m},\star }(\{2\}^a, 4, \{2\}^b) $ by first applying the stuffle antipode to obtain an expression involving only $ \zeta ^{\mathfrak {m}}(\{2\}^j, 4, \{2\}^i) $ , which has a simpler coaction.

We see that only the following subsequences can contribute to the motivic coaction of $ \operatorname {\mathrm {D}}_{2r+1} \zeta ^{\mathfrak {m}}(\{2\}^a, 4, \{2\}^b) $ ; all other subsequences will start and end at letters of the same parity

Hence, we have

$$ \begin{align*} & \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}(\{2\}^a, 4, \{2\}^b)\\ &\quad = - \delta_{r \leq a} \zeta^{\mathfrak{l}}_1(\{2\}^r) \otimes \zeta^{\mathfrak{m}}(\{2\}^{a-r}, 3, \{2\}^b) + \delta_{r \leq b} \zeta^{\mathfrak{l}}_1(\{2\}^r) \otimes \zeta^{\mathfrak{m}}(\{2\}^a, 3, \{2\}^{b-r}) . \end{align*} $$

Now, with the stuffle antipode formula extracted from Equation (A.2), we compute

$$ \begin{align*} & D_{2r+1 }\zeta^{\mathfrak{m},\star}(\{2\}^a, 4, \{2\}^b) \\ &\quad = \sum_{i=0}^a \sum_{j=0}^b (-1)^{i+j} \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m}}(\{2\}^j, 4, \{2\}^i) \Big( 1 \otimes \zeta^{\mathfrak{m},\star}(\{2\}^{a-i}) \zeta^{\mathfrak{m},\star}(\{2\}^{b-j}) \Big) \\ &\quad = \zeta^{\mathfrak{l}}_1(\{2\}^r) \otimes \sum_{i=0}^a \sum_{j=0}^b (-1)^{i+j} \big( - \delta_{r \leq j}\zeta^{\mathfrak{m}}(\{2\}^{j-r}, 3, \{2\}^i) + \delta_{r \leq i} \zeta^{\mathfrak{m}}(\{2\}^j, 3, \{2\}^{i-r}) \big) \\ & \qquad{} \cdot \zeta^{\mathfrak{m},\star}(\{2\}^{a-i}) \zeta^{\mathfrak{m},\star}(\{2\}^{b-j}). \end{align*} $$

Here, we can apply the motivic evaluation of $ \zeta ^{\mathfrak {m}}(\{2\}^\alpha , 3, \{2\}^\beta ) $ established by Brown [Reference Brown5]. Alternatively, we can apply the stuffle antipode again to rewrite the result instead of involving $ \zeta ^{\mathfrak {m},\star }(\{2\}^\alpha , 3, \{2\}^\beta ) $ and appeal to the motivic evaluation thereof, for a more direct formula (Glanois [Reference Glanois18] claims that the motivic evaluation of $ \zeta ^{\mathfrak {m},\star }(\{2\}^\alpha , 3, \{2\}^\beta ) $ requires knowing exactly certain conjectural identities among so-called $ \zeta ^{\star \star } $ values, however, it seems that the stuffle antipode formula allows one to automatically transfer the $ \zeta ^{\mathfrak {m}}(\{2\}^\alpha , 3, \{2\}^\beta ) $ evaluation to a corresponding $ \zeta ^{\mathfrak {m},\star }(\{2\}^\beta , 3, \{2\}^\alpha ) $ evaluation).

After (separately) shifting $ i $ and $ j $ by $ r $ , which gives the factor $ (-1)^r $ below (and taking care with the signs; use the correspondence $ j \leftrightarrow b $ , $ i \rightarrow a $ ), we find

$$ \begin{align*} & \operatorname{\mathrm{D}}_{2r+1 }\zeta^{\mathfrak{m},\star}(\{2\}^a, 4, \{2\}^b) \\ &\quad = (-1)^{r} \zeta^{\mathfrak{l}}_1(\{2\}^r) \otimes \big( \delta_{r\leq a} \zeta^{\mathfrak{m},\star}(\{2\}^{a-r}, 3, \{2\}^{b}) - \delta_{r \leq b} \zeta^{\mathfrak{m},\star}(\{2\}^a, 3,\{2\}^{b-r}) \big) . \end{align*} $$

We note that this is essentially the same expression as one obtains with Glanois’s setup involving the motivic coaction on $ \zeta ^{\star } $ values, after applying the dihedral symmetries to simplify terms in the coalgebra on the left-hand side. One only needs to apply the result that $ (-1)^r \zeta ^{\mathfrak {l}}_1(\{2\}^r) = 2 \zeta ^{\mathfrak {l}}(2r+1) = -\zeta ^{\mathfrak {l},\star }_1(\{2\}^r) $ to obtain exactly the same formula.

Now, apply the following motivic evaluations

$$ \begin{align*} \zeta^{\mathfrak{l}}_1(\{2\}^r) & = 2 (-1)^r \zeta^{\mathfrak{l}}(2r+1) \\[3pt] \zeta^{\mathfrak{m},\star}(\{2\}^a, 3, \{2\}^b) & = - 2 \sum_{s=1}^{a+b+1} \bigg[ \binom{2s}{2a} {-} \delta_{s=a} - (1 - 2^{-2s}) \binom{2s}{2b+1} \bigg] \zeta^{\star,\mathfrak{m}}(\{2\}^{a+b+1-s}) \zeta^{\mathfrak{m}}(2s+1) \end{align*} $$

along with $ \zeta ^{\star ,\mathfrak {m}}(\{2\}^n) = -2 \zeta ^{\mathfrak {m}}(\overline {2n}) . $ We find

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} \zeta^{\mathfrak{m},\star}(\{2\}^a, 4, \{2\}^b) &= 8 \zeta^{\mathfrak{l}}(2r+1) \otimes \sum_{s=1}^{a+b+1-r} \bigg[ \binom{2s}{2a-2r} - \delta_{s=a-r} - (1-2^{-2s}) \binom{2s}{2b+1} \bigg] \\[3pt] & \quad\cdot \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2s - 2r}) \zeta^{\mathfrak{m}}(2s+1) \\[3pt] &\quad - 8 \zeta^{\mathfrak{l}}(2r+1) \otimes \sum_{s=1}^{a+b+1-r} \bigg[ \binom{2s}{2a} - \delta_{s=a} - (1-2^{-2s}) \binom{2s}{2b-2r+1} \bigg] \\[3pt] &\quad \cdot \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2s - 2r}) \zeta^{\mathfrak{m}}(2s+1) . \end{align*} $$

Comparison of left- and right-hand side of Equation (B.3)

Firstly, make the change of variables $ s \mapsto a + b + 1 - s - r $ in the sums for $ \operatorname {\mathrm {D}}_{2r+1} ( \text {RHS Equation}~({\textrm{B.3}}) ) $ ; after considering the cases in each resulting delta term – and dropping terms $ \zeta ^{\mathfrak {m}}(1) = 0 $ by regularisation– we find

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} (\text{RHS Equation}~({\textrm{B.3}})) & = 8 \delta_{r=a} \zeta^{\mathfrak{l}}(2r+1) \otimes \sum_{k=1}^{b+1} \zeta^{\mathfrak{m}}(2k+1) \zeta^{\mathfrak{m}}(\overline{2b+2-2k}) \\ &\quad + 8 \delta_{r \leq b+1} \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta^{\mathfrak{m}}(2a+1) \zeta^{\mathfrak{m}}(\overline{2b + 2 - 2r}) \\ &\quad - 8 \sum_{s = \max(1, b-r+1)}^{a+b+1-r} \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( {-} \binom{2s}{2b+1} + \binom{2s}{2b+1-2r} \bigg) \\ &\quad \cdot \zeta^{\mathfrak{m}}(\overline{2s+1}) \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2r - 2s)}) \\ &\quad - 8 \sum_{s=\max(1, a-r+1)}^{a+b+1-r} \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( \binom{2s}{2a} + \delta_{r=a} - \binom{2s}{2a-2r} \bigg) \\ &\quad \cdot \zeta^{\mathfrak{m}}(2s + 1) \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2r - 2s)}) . \end{align*} $$

Note, here, that the two terms involving $ \delta _{r=a} $ cancel. Then the sums over $ s $ may be extended to start at $ s = 1 $ . The first sum needs no correction term, as the numerators of each binomial are strictly greater than the denominators in this case, however, the term $ \binom {2s}{2a-2r} $ in the second sum needs to be corrected when $ s = a-r $ for $ a-r \geq 1 $ . We obtain

$$ \begin{align*} \operatorname{\mathrm{D}}_{2r+1} (\text{RHS Equation}~({\textrm{B.3}})) &= 8 \delta_{r \leq b+1} \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta^{\mathfrak{m}}(2a+1) \zeta^{\mathfrak{m}}(\overline{2b + 2 - 2r}) \\ &\quad - 8 \delta_{r \leq a-1} \zeta^{\mathfrak{l}}(2r+1) \otimes \zeta^{\mathfrak{m}}(2a + 1 - 2r) \zeta^{\mathfrak{m}}(\overline{2b+2}) \\ &\quad - 8 \sum_{s = 1}^{a+b+1-r} \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( {-} \binom{2s}{2b+1} + \binom{2s}{2b+1-2r} \bigg) \\ &\quad \cdot \zeta^{\mathfrak{m}}(\overline{2s+1}) \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2r - 2s)}) \\ &\quad - 8 \sum_{s=1}^{a+b+1-r} \zeta^{\mathfrak{l}}(2r+1) \otimes \bigg( \binom{2s}{2a} - \binom{2s}{2a-2r} \bigg) \\ &\quad \cdot \zeta^{\mathfrak{m}}(2s + 1) \zeta^{\mathfrak{m}}(\overline{2a + 2b + 2 - 2r - 2s)}) .\end{align*} $$

Finally, write $ \zeta ^{\mathfrak {m}}(\overline {2s+1}) = -(1-2^{2s})\zeta ^{\mathfrak {m}}(2s+1) $ . It is now straightforward to check that $ \operatorname {\mathrm {D}}_{2r+1} ( \text {LHS Equation}~({\textrm{B.3}}) ) = \operatorname {\mathrm {D}}_{2r+1} ( \text {RHS Equation}~({\textrm{B.3}}) ) $ ; the two terms outside the sum for $ \operatorname {\mathrm {D}}_{2r+1} ( \text {RHS Equation}~({\textrm{B.3}}) ) $ above correspond to the deltas terms in the expression for $ \operatorname {\mathrm {D}}_{2r+1}( \text {LHS Equation}~({\textrm{B.3}}) ) $ .

This completes the proof of Proposition B.3, and shows the reduction of $ \zeta ^{\star }(\{2\}^a, 4, \{2\}^b) $ to depth 3 alternating MZVs is motivic.

B.3 Motivic evaluation of $ \zeta ^{\mathfrak {l}}(\{2\}^a, 4, \{2\}^b) $

Now that we have verified all of the ingredients for the evaluations of $ \zeta (\{2\}^a, 4, \{2\}^b) $ and $ \zeta ^\star (\{2\}^a, 4, \{2\}^b) $ are motivic, we may conclude that the identities in Theorems A.6 and A.7 hold for $ \zeta ^{(\star )} $ replaced by their motivic counterparts, and $ \mathrm {i}\pi $ replaced by $ \frac {1}{2} \mathbb {L}^{\mathfrak {m}} = (\mathrm {i}\pi )^{\mathfrak {m}} $ .

More importantly, the evaluation of $ \zeta (\{2\}^a, 4, \{2\}^b) $ modulo products from Corollary A.8 is also motivic, and we obtain the result of Lemma 4.1 as an immediate corollary.

Corollary B.4. The following evaluation holds in the motivic coalgebra