2.1. Connections and curvatures

In this subsection, we recall the definitions of the Chern connection and its curvature for a Hermitian holomorphic vector bundle. One can refer to [Reference Kobayashi16, Chapter 1] for more details. We use the Einstein summation convention in this paper.

Let $\pi :(E,h^E)\to X$ be a Hermitian holomorphic vector bundle over a complex manifold X, $\mathrm {rank} E=r$ and $\dim X=n$ . Let $\nabla ^E$ be the Chern connection of $(E,h^E)$ , which preserves the metric $h^E$ and is of $(1,0)$ -type. With respect to a local holomorphic frame $\{e_i\}_{1\leq i\leq r}$ of E, one has

(2.1) $$ \begin{align} \nabla^E e_i=\theta^j_i e_j, \end{align} $$

where $\theta =(\theta ^j_i)$ (j row, i column) is the connection form of $\nabla ^E$ . More precisely,

$$ \begin{align*} \theta^j_i=\partial h_{i\bar{k}}h^{\bar{k}j}, \end{align*} $$

where $h_{i\bar {k}}:=h(e_i,{e_k})$ . In terms of matrix form, it is

$$ \begin{align*} \theta^\top=\partial h\cdot h^{-1}. \end{align*} $$

Considering $e=(e_1,\cdots ,e_r)$ as a row vector, then (2.1) can be written as

$$ \begin{align*} \nabla^E e=e\cdot\theta. \end{align*} $$

Let $R^E=(\nabla ^E)^2\in A^{1,1}(X,\mathrm {End}(E))$ denote the Chern curvature of $(E,h^E)$ , and write

$$ \begin{align*} R^E=R^j_ie_j\otimes e^i \in A^{1,1}(X,\mathrm{End}(E)), \end{align*} $$

where $R=(R^j_i)$ (j row, i column) is the curvature matrix whose entries are $(1,1)$ -forms and $\{e^i\}_{1\leq i\leq r}$ denotes the dual frame of $\{e_i\}_{1\leq i\leq r}$ . The curvature matrix $R=(R^j_i)$ is given by

$$ \begin{align*} R^{j}_i=d\theta^j_i+\theta^j_k\wedge \theta^k_i=\bar{\partial}\theta^j_i. \end{align*} $$

If $\{\tilde {e}_i\}_{1\leq i\leq r}$ is another local holomorphic frame of E with $\tilde {e}_i=a_i^je_j$ , then

$$ \begin{align*} \tilde{e}=e\cdot a, \end{align*} $$

where $\tilde {e}:=(\tilde {e}_1,\cdots ,\tilde {e}_r)$ and $a=(a^i_j)$ . Denote by $\widetilde {R}$ the curvature matrix with respect to the local frame $\{\tilde {e}_i\}_{1\leq i\leq r}$ . Then

(2.2) $$ \begin{align} \widetilde{R}=a^{-1}\cdot R\cdot a. \end{align} $$

Let $\{z^\alpha \}_{1\leq \alpha \leq n}$ be local holomorphic coordinates of X. Write

$$ \begin{align*} R^j_i=R^j_{i\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta \end{align*} $$

and denote

$$ \begin{align*} R_{i\bar{j}}:=R^{k}_{i}h_{k\bar{j}}=R_{i\bar{j}\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta, \end{align*} $$

so that

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}=R^k_{i\alpha\bar{\beta}}h_{k\bar{j}}=-\partial_\alpha\partial_{\bar{\beta}}h_{i\bar{j}}+h^{\bar{l}k}\partial_\alpha h_{i\bar{l}}\partial_{\bar{\beta}}h_{k\bar{j}}, \end{align*} $$

where $\partial _\alpha :=\partial /\partial z^\alpha $ and $\partial _{\bar {\beta }}:=\partial /\partial \bar {z}^\beta $ .

2.2. Strongly decomposable positivity

This subsection defines two types of strongly decomposably positive vector bundles. Firstly, we recall the following definitions of Nakano positive and dual Nakano positive vector bundles.

Definition 2.1 ((Dual) Nakano positive)

A Hermitian holomorphic vector bundle $(E,h^E)$ is called Nakano positive (resp. non-negative) if

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{u^{j\beta}}>0 \,(\text{resp. } \geq 0 ) \end{align*} $$

for any non-zero element $u=u^{i\alpha }e_i\otimes \partial _\alpha \in E\otimes T^{1,0}X$ . $(E,h^E)$ is called dual Nakano positive (non-negative) if

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}v^{\bar{j}\alpha}\overline{v^{\bar{i}\beta}}>0 \,(\text{resp. } \geq 0 ) \end{align*} $$

for any non-zero element $v=v^{\bar {j}\alpha }\bar {e}_j\otimes \partial _\alpha \in \overline {E}\otimes T^{1,0}X$ .

In [Reference Finski9, Definition 2.18], S. Finski introduced the following new notion of positivity for vector bundles: decomposable positivity.

Definition 2.2 (Decomposably positive)

A Hermitian vector bundle $\left(E, h^E\right)$ is called decomposably non-negative if for any $x \in X$ , there is a number $N \in \mathbb {N}$ and linear (respectively sesquilinear) forms $l_p^{\prime }: T_x^{1,0} X \otimes E_x \rightarrow \mathbb {C}$ (respectively $l_p: T_x^{1,0} X \otimes E_x \rightarrow \mathbb {C}$ ), $p=1, \ldots , N$ , such that for any $v \in T_x^{1,0} X, \xi \in E_x$ , we have

$$ \begin{align*}\frac{1}{2 \pi}\left\langle R_x^E(v, \bar{v}) \xi, \xi\right\rangle_{h^E}=\sum_{p=1}^N\left|l_p(v, \xi)\right|^2+\sum_{p=1}^N\left|l_p^{\prime}(v, \xi)\right|^2. \end{align*} $$

We say that it is decomposably positive if, moreover, $\left\langle R_x^E(v, \bar {v}) \xi , \xi \right\rangle _{h^E} \neq 0$ for $v, \xi \neq 0$ .

Remark 2.3. From the above definition, a (dual) Nakano positive vector bundle must be decomposably positive. From [Reference Finski9, Proposition 2.21], for $n\cdot r\leq 6$ decomposable positivity is equivalent to Griffiths positivity; that is,

(2.3) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}v^i\bar{v}^j\xi^\alpha\bar{\xi}^\beta>0 \end{align} $$

for any non-zero $\xi =\xi ^\alpha \partial _\alpha \in T^{1,0}X$ and $v=v^ie_i\in E$ . Decomposable positivity is strictly stronger than Griffiths positivity for all other $n, r \neq 1$ .

Next, we introduce the notion of strongly decomposable positivity, which falls in between (dual) Nakano positivity and decomposable positivity.

Definition 2.4 (Strongly decomposably positive of type I)

A Hermitian vector bundle $(E,h^E)$ is called a strongly decomposably positive (resp. non-negative) vector bundle of type I if, for any $x\in X$ , there exists a decomposition $T^{1,0}_x X=U_x\oplus V_x$ such that

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{u^{j\beta}}>0(\text{resp. }\geq 0), \,\, R_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{v{^{\prime}j\beta}}=0,\,\, R_{i\bar{j}\alpha\bar{\beta}}v^{\bar{j}\alpha}\overline{v^{\bar{i}\beta}}>0(\text{resp. }\geq 0) \end{align*} $$

for any non-zero elements $u=u^{i\alpha }e_i\otimes \partial _\alpha \in E_x\otimes U_x$ , $v'=v{^{\prime }j\beta }e_j\otimes \partial _\beta \in E_x\otimes V_x$ and $v=v^{\bar {j}\alpha }\bar {e}_j\otimes \partial _\alpha \in \overline {E}_x\otimes V_x$ .

For any point $x\in X$ , if $V_x=\{0\}$ (resp. $U_x=\{0\}$ ), then strongly decomposable positivity of type I is exactly Nakano positive (resp. dual Nakano positive).

Example 2.6. Let $\pi _1:(E_1,h^{E_1})\to X_1$ be a Nakano positive vector bundle and $\pi _2:(E_2,h^{E_2})\to X_2$ be a dual Nakano positive vector bundle. Denote by $p_i:X_1\times X_2\to X_i$ , $i=1,2$ , the natural projections. Then

$$ \begin{align*} \pi_{\oplus}:(p_1^*E_1\oplus p_2^*E_2,p_1^*h^{E_1}\oplus p_2^*h^{E_2})\to X_1\times X_2 \end{align*} $$

is a strongly decomposably non-negative vector bundle of type I, and

$$ \begin{align*} \pi_{\otimes}:(p_1^*E_1\otimes p_2^*E_2,p_1^*h^{E_1}\otimes p_2^*h^{E_2})\to X_1\times X_2 \end{align*} $$

is a strongly decomposably positive vector bundle of type I.

From Definition 2.4, a strongly decomposably positive vector bundle $(E,h^E)$ of type I means that there is a decomposition of holomorphic tangent bundle $T^{1,0}_x X=U_x\oplus V_x$ , such that $(E,h^E)$ is Nakano positive in $E_x\otimes U_x$ and dual Nakano positive in $\overline {E}_x\otimes V_x$ , and the cross curvature terms vanish. Naturally, one may define another strongly decomposable positivity of vector bundles by decomposing the vector bundle.

Definition 2.7 (Strongly decomposably positive of type II)

We call $(E,h^E)$ a strongly decomposably positive (resp. non-negative) vector bundle of type II if, for any $x\in X$ , there is an orthogonal decomposition of $(E_x,h^{E}|_{E_x})$ , $E_x=E_{1,x}\oplus E_{2,x}$ , such that the Chern curvature $R^E_x$ has the form

$$ \begin{align*} R^E_x=\begin{pmatrix} R^{E}_x|_{E_{1,x}}&0 \\ 0& R^{E}_x|_{E_{2,x}} \end{pmatrix}, \end{align*} $$

and $R_{i\bar {j}\alpha \bar {\beta }}u^{i\alpha }\overline {u^{j\beta }}>0$ (resp. $\geq 0$ ) for any non-zero $u=u^{i\alpha } e_i\otimes \partial _\alpha \in E_{1,x}\otimes T^{1,0}_xX$ , $R_{i\bar {j}\alpha \bar {\beta }}v^{i\bar {\beta }}\overline {v^{j\bar {\alpha }}}>0$ (resp. $\geq 0$ ) for any non-zero $v=v^{i\bar {\beta }}e_i\otimes \partial _{\bar {\beta }}\in E_{2,x}\otimes T^{0,1}_xX$ .

A simple example of a strongly decomposably vector bundle of type II is as follows.

By Definition 2.4 and Definition 2.7, a strongly decomposably positive vector bundle is defined as follows.

Similarly, one can define strongly decomposably negative (non-positive) vector bundles. Note that the dual of the Nakano positive (negative) vector bundle is dual Nakano negative (positive), so we have the following.

Remark 2.11. Let $(E,h^E)$ be a strongly decomposably positive vector bundle and Q be a quotient bundle of E. The curvature of the bundle Q is given by

$$ \begin{align*} R^Q=\left.R^E\right|_Q+C \wedge \overline{C}^\top \end{align*} $$

for some matrix C whose entries are $(1,0)$ -forms. From the criteria of strongly decomposably positive vector bundles, Theorem 4.3 and Theorem 5.2, and the above curvature formula of quotient bundles, the quotient bundle $(Q,h^Q)$ ceases to be a strongly decomposably positive vector bundle in general.

2.3. Relation to decomposable positivity

From the equivalent descriptions of Nakano non-negative and dual non-negative due to S. Finski [Reference Finski9, Theorem 2.15 and 2.17], one has the following:

Proposition 2.12. A Hermitian vector bundle $(E,h^E)$ is decomposably non-negative if and only if the Chern curvature matrix has the form

(2.4) $$ \begin{align} R=-B\wedge\overline{B}^\top+A\wedge\overline{A}^\top \end{align} $$

with respect to a unitary frame, where A (resp. B) is a $r\times N$ matrix with $(1,0)$ -forms (resp. $(0,1)$ -forms) as entries.

Proof. From Definition 2.2, $(E,h^E)$ is decomposably positive if for any $x \in X$ , there is a number $N \in \mathbb {N}$ and linear (respectively sesquilinear) forms $l_p^{\prime }: T_x^{1,0} X \otimes E_x \rightarrow \mathbb {C}$ (respectively $l_p: T_x^{1,0} X \otimes E_x \rightarrow \mathbb {C}$ ), $p=1, \ldots , N$ , such that for any $v \in T_x^{1,0} X, \xi \in E_x$ , we have

(2.5) $$ \begin{align} \left\langle R_x^E(v, \bar{v}) \xi, \xi\right\rangle_{h^E}=\sum_{p=1}^N\left|l_p(v, \xi)\right|^2+\sum_{p=1}^N\left|l_p^{\prime}(v, \xi)\right|^2. \end{align} $$

We denote

$$ \begin{align*} l_p(v,\xi)=l_{ip\bar{\beta}}v^i\bar{\xi}^\beta,\quad l^{\prime}_p(v,\xi)=l^{\prime}_{ip\alpha}v^i\xi^\alpha, \end{align*} $$

and set $A=(A_{jp})$ and $B=(B_{jp})$ by

$$ \begin{align*} A_{jp}:=A_{jp\alpha}dz^\alpha=\overline{l_{jp\bar{\alpha}}}dz^\alpha,\quad B_{jp}:=B_{jp\bar{\beta}}d\bar{z}^\beta=\overline{l^{\prime}_{jp\beta}}d\bar{z}^\beta. \end{align*} $$

Then (2.5) is equivalent to

(2.6) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}&=\sum_{p=1}^N l_{ip\bar{\beta}}\overline{l_{jp\bar{\alpha}}}+\sum_{p=1}^N l^{\prime}_{ip\alpha}\overline{l^{\prime}_{jp\beta}} \nonumber\\ &=\sum_{p=1}^N A_{jp\alpha}\overline{A_{ip\beta}}+\sum_{i=1}^N B_{jp\bar{\beta}}\overline{B_{ip\bar{\alpha}}}. \end{align} $$

With respect to a unitary frame, $R^j_{i\alpha \bar {\beta }}=R_{i\bar {j}\alpha \bar {\beta }}$ and (2.6) is equivalent to

$$ \begin{align*} R=-B\wedge\overline{B}^\top+A\wedge\overline{A}^\top, \end{align*} $$

which completes the proof.

From Proposition 2.12 and Definition 2.2, $(E,h^E)$ is decomposably positive if and only if (2.4) holds and $\left\langle R_x^E(v, \bar {v}) \xi , \xi \right\rangle _{h^E} \neq 0$ for $v, \xi \neq 0$ . By Theorem 4.3 and Theorem 5.2, we have the following:

However, from Theorem 4.3 and Theorem 5.2, the two types of strongly decomposably positive vector bundles cannot contain each other. Both are the generalizations of (dual) Nakano positive vector bundles and are stronger than decomposable positivity. One can refer to the following Figure 1.

Figure 1 Relations of several notions of positivity.

4.1. A criterion of type I positivity

In this subsection, following S. Finski’s approach [Reference Finski9, Theorem 2.15, 2.17], we give a criterion for the strongly decomposable positivity (non-negativity) of type I by using M.-D. Choi’s results.

Let $(E,h^E)$ be a strongly decomposably non-negative vector bundle of type I. For any $x\in X$ , there exists a decomposition $T^{1,0}_xX=U_x\oplus V_x$ . One can take local holomorphic coordinates $\{z^1,\cdots , z^n\}$ around x such that

$$ \begin{align*} U_x=\mathrm{span}_{\mathbb{C}}\{\partial_1,\cdots, \partial_{n_0}\},\,\, V_x=\mathrm{span}_{\mathbb{C}}\{\partial_{n_0+1},\cdots, \partial_{n}\}, \end{align*} $$

where $n_0:=\dim U_x$ and recall that $\partial _\alpha :=\partial /\partial z^\alpha $ . Let $\{e_i\}_{1\leq i\leq r}$ be a local holomorphic frame of E such that

$$ \begin{align*} h_{i\bar{j}}(x)=h^E(e_i(x),e_j(x))=\delta_{ij}. \end{align*} $$

With respect to $\{z^\alpha \}_{1\leq \alpha \leq n}$ and $\{e_i\}_{1\leq i\leq r}$ , the Chern curvature matrix $R=(R^j_i)$ at $x\in X$ has the following expression:

(4.1) $$ \begin{align} R^j_i&=R^j_{i\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta \nonumber\\ &=\sum_{\alpha,\beta=1}^{n_0}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta+\sum_{\alpha,\beta=n_0+1}^{n}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta\\ &\quad+\sum_{\alpha=1}^{n_0}\sum_{\beta=n_0+1}^{n}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta+\sum_{\alpha=n_0+1}^{n}\sum_{\beta=1}^{n_0}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta.\nonumber \end{align} $$

By assumption, $(E,h^E)$ is strongly decomposably non-negative of type I, so

$$ \begin{align*} \sum_{\alpha=1}^{n_0}\sum_{\beta=n_0+1}^nR_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{v{^{\prime}j\beta}}=0 \end{align*} $$

for any $\sum _{\alpha =1}^{n_0}u^{i\alpha }e_i\otimes \partial _\alpha $ and $\sum _{\beta ={n_0+1}}^nv{^{\prime }j\beta }e_j\otimes \partial _\beta $ , which follows that

(4.2) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}=0,\quad 1\leq \alpha\leq n_0,\,n_0+1\leq \beta\leq n. \end{align} $$

By conjugation, one gets

(4.3) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}=\overline{R_{j\bar{i}\beta\bar{\alpha}}}=0,\quad n_0+1\leq \alpha\leq n,\,1\leq \beta\leq n_0. \end{align} $$

Substituting (4.2), (4.3) into (4.1), one has

$$ \begin{align*} R^j_i=\sum_{\alpha,\beta=1}^{n_0}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta+\sum_{\alpha,\beta=n_0+1}^{n}R_{i\alpha\bar{\beta}}^j dz^\alpha\wedge d\bar{z}^\beta. \end{align*} $$

For the local frame $\{\partial _\alpha \}_{1\leq \alpha \leq n}$ , we define a local metric g around x by

$$ \begin{align*} g_{\alpha\bar{\beta}}= g(\partial_\alpha,\partial_\beta):=\delta_{\alpha\beta}. \end{align*} $$

Now we define a linear map

$$ \begin{align*} H^V_x: \mathrm{End}(V_x)\to \mathrm{End}(E_x) \end{align*} $$

by

(4.4) $$ \begin{align} H_x^V(\partial_\alpha\otimes dz^\gamma)=R^j_{i \alpha \bar{\beta}} g^{\bar{\beta} \gamma} e_j \otimes e^i=R^j_{i \alpha \bar{\gamma}} e_j \otimes e^i \end{align} $$

for any $n_0+1\leq \alpha ,\gamma \leq n$ . With respect to the basis $\{\partial _\alpha \}_{n_0+1\leq \alpha \leq n}$ , the matrix of $\partial _\alpha \otimes dz^\gamma \in \mathrm {End}(U_x)$ is $E_{\alpha \gamma }$ , which is the $(n-n_0)\times (n-n_0)$ matrix with $1$ at the $(\alpha ,\gamma )$ -component and zeros elsewhere. The matrix of $R^j_{i \alpha \bar {\gamma }} e_j \otimes e^i\in \mathrm {End}(E_x)$ is given by $(R^j_{i \alpha \bar {\gamma }})_{1\leq j,i\leq r}$ (j row, i column). In terms of matrices, (4.4) becomes

(4.5) $$ \begin{align} H^V_x(E_{\alpha\gamma})=(R^j_{i \alpha \bar{\gamma}})_{1\leq j,i\leq r}=(R_{i \bar{j} \alpha \bar{\gamma}})_{1\leq j,i\leq r}. \end{align} $$

Then $(H^V_x(E_{\alpha \gamma }))_{n_0+1\leq \alpha ,\gamma \leq n}$ is a $(n-n_0)\times (n-n_0)$ block matrix with $r\times r$ matrices as entries, and

$$ \begin{align*} &(H^V_x(E_{\alpha\gamma}))_{n_0+1\leq \alpha,\gamma\leq n}=\left((R_{i \bar{j} \alpha \bar{\gamma}})_{1\leq j,i\leq r}\right)_{n_0+1\leq \alpha,\gamma\leq n} \quad (j\, \alpha\text{ row}, i\,\gamma\text{ column}). \end{align*} $$

Since $(E,h^E)$ is strongly decomposably non-negative of type I, then

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}v^{\bar{j}\alpha}\overline{v^{\bar{i}\beta}}\geq 0 \end{align*} $$

for any non-zero $v=v^{\bar {j}\alpha }e_i\otimes \partial _\alpha \in E_x\otimes V_x$ , which follows that the matrix

$$ \begin{align*}(H^V_x(E_{\alpha\gamma}))_{n_0+1\leq \alpha,\gamma\leq n}\end{align*} $$

is positive semi-definite. By using [Reference Choi4, Theorem 2 and Theorem 1], there exist $(n-n_0)\times r$ matrices $V_p, 1\leq p\leq N_1$ (one can choose $N_1=(n-n_0)\cdot r$ ) such that

$$ \begin{align*} H^V_x(E_{\alpha\gamma})=\sum_{p=1}^{N_1} \overline{V_p}^\top\cdot E_{\alpha\gamma}\cdot V_p \end{align*} $$

for any $n_0+1\leq \alpha ,\gamma \leq n$ . Combining with (4.5) and considering the $(j,i)$ entry, one has

$$ \begin{align*} R^j_{i\alpha\bar{\beta}}=\sum_{p=1}^{N_1} (\overline{V_p}^\top\cdot E_{\alpha\beta}\cdot V_p)_{j,i}=\sum_{p=1}^{N_1}\overline{(V_p)_{\alpha j}}(V_p)_{\beta i}. \end{align*} $$

Hence,

$$ \begin{align*} \sum_{\alpha,\beta=n_0+1}^{n}R^j_{i\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta &=\sum_{\alpha,\beta=n_0+1}^{n}\sum_{p=1}^{N_1}\overline{(V_p)_{\alpha j}}(V_p)_{\beta i}dz^\alpha\wedge d\bar{z}^\beta\\ &=\sum_{p=1}^{N_1}A_{jp}\wedge \overline{A_{i p}}, \end{align*} $$

where $A_{jp}:=\sum _{\alpha =n_0+1}^n\overline {(V_p)_{\alpha j}}dz^\alpha $ , and one has

$$ \begin{align*} \left( \sum_{\alpha,\beta=n_0+1}^{n}R^j_{i\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta\right)_{1\leq j,i\leq r}=A\wedge \overline{A}^\top, \end{align*} $$

where $A=(A_{jp})$ is a $r\times N_1$ matrix with $(1,0)$ -forms in $V^*_x$ as entries.

Similarly, by considering the linear map

$$ \begin{align*} H^U_x:\mathrm{End}(U_x)\to \mathrm{End}(E_x^*), \quad H_x^V(\partial_\alpha\otimes dz^\gamma)=R^j_{i \alpha \bar{\gamma}} e^i\otimes e_j. \end{align*} $$

One can obtain that

$$ \begin{align*} &(H^V_x(E_{\alpha\gamma}))_{1\leq \alpha,\gamma\leq n_0}=\left((R_{i \bar{j} \alpha \bar{\gamma}})_{1\leq j,i\leq r}\right)_{1\leq \alpha,\gamma\leq n_0} \quad (i\, \alpha\text{ row}, j\,\gamma\text{ column}), \end{align*} $$

which follows that

$$ \begin{align*} \left( \sum_{\alpha,\beta=1}^{n_0}R^j_{i\alpha\bar{\beta}}dz^\alpha\wedge d\bar{z}^\beta\right)_{1\leq j,i\leq r}=-B\wedge \overline{B}^\top, \end{align*} $$

where $B=(B_{jp})$ is a $r\times N_2$ (one can choose $N_2=n_0\cdot r$ ) matrix with $(0,1)$ -forms in $\overline {U^*_x}$ as entries.

Thus, if $(E,h^E)$ is strongly decomposably non-negative of type I, then for any $x\in X$ , the Chern curvature matrix at this point has the form

(4.6) $$ \begin{align} R=-B\wedge \overline{B}^\top+A\wedge \overline{A}^\top \end{align} $$

with respect to a unitary frame, where B is a $r\times N_2$ matrix with $(0,1)$ -forms as entries, A is a $r\times N_1$ matrix with $(1,0)$ -forms as entries, and

(4.7) $$ \begin{align} \mathrm{span}_{\mathbb{C}}\{\overline{B}\}\cap \mathrm{span}_{\mathbb{C}}\{A\}=\{0\}, \end{align} $$

where

$$ \begin{align*}\{\overline{B}\}:=\{\overline{B_{ip}}, {1\leq i\leq r,1\leq p\leq N_2}\}\end{align*} $$

and

$$ \begin{align*}\{A\}:=\{A_{ip},1\leq i\leq r, 1\leq p\leq N_1\}.\end{align*} $$

Remark 4.1. It is noted that the above argument is independent of the choice of unitary frames. If $\tilde {e}=e\cdot a$ is also a unitary frame, then a is a unitary matrix. By (2.2), one has

$$ \begin{align*} \widetilde{R}&=a^{-1}\cdot(-B\wedge \overline{B}^\top+A\wedge \overline{A}^\top)\cdot a\\ &=-a^{-1}B\wedge \overline{a^{-1}B}^\top+a^{-1}A\wedge \overline{a^{-1}A}^\top, \end{align*} $$

which has the form (4.6). Moreover, $\mathrm {span}_{\mathbb {C}}\{\overline {B}\}=\mathrm {span}_{\mathbb {C}}\{\overline {a^{-1}B}\}$ and $ \mathrm {span}_{\mathbb {C}}\{A\}= \mathrm {span}_{\mathbb {C}}\{a^{-1}A\}$ , (4.7) is equivalent to

$$ \begin{align*} \mathrm{span}_{\mathbb{C}}\{\overline{a^{-1}B}\}\cap \mathrm{span}_{\mathbb{C}}\{a^{-1}A\}=\{0\}. \end{align*} $$

Conversely, we assume (4.6) and (4.7) hold. For any $x\in X$ , taking local holomorphic coordinates $\{z^\alpha \}_{1\leq \alpha \leq n}$ around $x\in X$ such that

$$ \begin{align*} \mathrm{span}_{\mathbb{C}}\{dz^1|_x,\cdots, dz^{n_0}|_x\}=\mathrm{span}_{\mathbb{C}}\{\overline{B}\} \end{align*} $$

and

$$ \begin{align*} \mathrm{span}_{\mathbb{C}}\{dz^{n_0+1}|_x,\cdots, dz^{n_1}|_x\}=\mathrm{span}_{\mathbb{C}}\{A\}, \end{align*} $$

we now set

$$ \begin{align*} U_x:= \mathrm{span}_{\mathbb{C}}\{\partial_1|_x,\cdots, \partial_{n_0}|_x\},\quad V_x=\mathrm{span}_{\mathbb{C}}\{\partial_{n_0+1}|_x,\cdots, \partial_{n}|_x\}. \end{align*} $$

Then $U_x\oplus V_x=T^{1,0}_xX$ . Using (4.6) and (4.7), one can check that $(E,h^E)$ is strongly decomposably non-negative.

Hence, $(E,h^E)$ is strongly decomposably non-negative of type I if and only if the Chern curvature matrix of $(E,h^E)$ satisfies (4.6) and (4.7).

Next, we assume that $(E,h^E)$ is strongly decomposably positive of type I; that is, it is strongly decomposably non-negative of type I and

(4.8) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{u^{j\beta}}=0\Longrightarrow u^{i\alpha}=0,\,\text{ for all }1\leq i\leq r,\,1\leq \alpha\leq n_0 \end{align} $$

and

(4.9) $$ \begin{align} R_{i\bar{j}\alpha\bar{\beta}}v^{\bar{j}\alpha}\overline{v^{\bar{i}\beta}}=0\Longrightarrow v^{\bar{j}\alpha}=0,\,\text{ for all }1\leq j\leq r,\,n_0+1\leq \beta\leq n. \end{align} $$

By the equivalent description of strongly decomposably non-negative of type I (i.e., (4.6) and (4.7) hold), one has

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}u^{i\alpha}\overline{u^{j\beta}}=\sum_{p=1}^{N_2}|B_{ip\bar{\alpha}}\overline{u^{i\alpha}}|^2. \end{align*} $$

Definition 4.2. Let B be a $r\times N_2$ matrix with $(0,1)$ -forms as entries. We define the following $N_2\times rn_0$ -matrix $\mathbf {B}$ as

(4.10) $$ \begin{align} \mathbf{B}:=(B_{ip\bar{\alpha}})_{p,i\alpha}=\begin{pmatrix} B_{11\overline{1}}& B_{11\overline{2}}&\cdots& B_{r1\overline{n_0}} \\ B_{12\overline{1}}& B_{11\overline{2}}&\cdots& B_{r2\overline{n_0}}\\ \vdots&\vdots&\ddots&\vdots\\ B_{1N_2\overline{1}}& B_{1N_2\overline{2}}&\cdots& B_{rN_2\overline{n_0}} \end{pmatrix}_{N_2\times rn_0}. \end{align} $$

Similarly, if A is a $r\times N_1$ matrix with $(1,0)$ -forms as entries, we define

(4.11) $$ \begin{align} \mathbf{A}:=(A_{jp\alpha})_{p,j\alpha}=\begin{pmatrix} A_{11(n_0+1)}& A_{11(n_0+2)}&\cdots& A_{r1n} \\ A_{12(n_0+1)}& A_{11(n_0+2)}&\cdots& A_{r2n}\\ \vdots&\vdots&\ddots&\vdots\\ A_{1N_1(n_0+1)}& A_{1N_1(n_0+2)}&\cdots& A_{rN_1n} \end{pmatrix}_{N_1\times r(n-n_0)}. \end{align} $$

Hence, (4.8) is equivalent to the equation $\mathbf {B}x=0$ has only zero solution. This is also equivalent to $\mathrm {rank}(\mathbf {B})=r n_0$ . Similarly, (4.9) is equivalent to $\mathrm {rank}(\mathbf {A})=r(n-n_0)$ .

We obtain the following.

As a result, we obtain the following criteria of (dual) Nakano positive vector bundles.

4.2. Weak positivity of Schur forms

In this subsection, we prove the weak positivity of Schur forms for strongly decomposably positive vector bundles of type I.

4.2.1. Schur polynomial

Each partition $\lambda \in \Lambda (k, r)$ gives rise to a Schur polynomial $P_\lambda \in \mathbb {Q}\left[c_1, \ldots , c_r\right]$ of degree k, defined as $k\times k$ determinant

$$ \begin{align*} P_\lambda\left(c_1, \ldots, c_r\right)=\operatorname{det}\left(c_{\lambda_i-i+j}\right)_{1 \leqslant i, j \leqslant k}, \end{align*} $$

where by convention $c_0=1$ and $c_i=0$ if $i \notin [0, r]$ .

Denote by $\mathrm {M}_r(\mathbb {C})$ and $\mathrm {G L}_r(\mathbb {C})$ the vector spaces of $r \times r$ matrices and the general linear group of degree r. A map $P: \mathrm {M}_r(\mathbb {C}) \rightarrow \mathbb {C}$ is called $\mathrm {G L}_r(\mathbb {C})$ -invariant if it is invariant under the conjugate action of $\mathrm {G L}_r(\mathbb {C})$ on $\mathrm {M}_r(\mathbb {C})$ . Now we define the following $\mathrm {G L}_r(\mathbb {C})$ -invariant function $c_i: \mathrm {M}_r(\mathbb {C}) \rightarrow \mathbb {C}, i=1, \ldots , r$ by

$$ \begin{align*}\operatorname{det}\left(I_r+t X\right)=\sum_{i=0}^r t^i \cdot c_i(X), \end{align*} $$

where $I_r$ is the identity matrix in $\mathrm {M}_r(\mathbb {C})$ . Then the graded ring of $\mathrm {G L}_r(\mathbb {C})$ -invariant homogeneous polynomials on $\mathrm {M}_r(\mathbb {C})$ , which we denote here by $\mathrm {I}(r)=\bigoplus _{k=0}^{+\infty } \mathrm {I}(r)_k$ , is multiplicatively generated by $c_1, \ldots , c_r$ .

Let $(E,h^E)$ be a Hermitian vector bundle; the i-th Chern form $c_i(E,h^E)$ is defined by

$$ \begin{align*}c_i\left(E, h^E\right)=c_i\left(\frac{\sqrt{-1} }{2 \pi}R^E\right). \end{align*} $$

For each $\lambda \in \Lambda (k,r)$ , recall the Schur form (see Introduction) can be given by

$$ \begin{align*} P_\lambda(c(E,h^E))=P_\lambda(c_1(E,h^E),\ldots, c_r(E,h^E)), \end{align*} $$

which represents the Schur class

$$ \begin{align*} P_\lambda(c(E)):=P_{\lambda}(c_1(E),\ldots, c_r(E))\in \mathrm{H}^{2k}(X,\mathbb{Z}). \end{align*} $$

4.2.2. Griffiths cone

By [Reference Griffiths14, Page 242, (5.6)], each $P\in \mathrm {I}(r)_k$ can be written as

(4.12) $$ \begin{align} P(B)=\sum_{\sigma, \tau \in S_k} \sum_{\rho \in[1, r]^k} p_{\rho \sigma \tau} B_{\rho_{\sigma(1)} \rho_{\tau(1)}} \cdots B_{\rho_{\sigma(k)} \rho_{\tau(k)}}, \end{align} $$

where $B_{\lambda \mu }, \lambda , \mu =1, \ldots , r$ are the components of the matrix B, $S_k$ is the permutation group on k indices and $[1, r]:=\{1, \ldots , r\}$ . An element $P\in \mathrm {I}(r)_k$ is called Griffiths non-negative if it can be expressed in the form (4.12) with

$$ \begin{align*}p_{\rho \sigma \tau}=\sum_{t \in T} \lambda_{\rho t} \cdot q_{\rho \sigma t} \bar{q}_{\rho \tau t} \end{align*} $$

for some finite set T, some real numbers $\lambda _{\rho t} \geqslant 0$ , and complex numbers $q_{\rho \sigma t}$ .

The Griffiths cone $\Pi (r) \subset \mathrm {I}(r)$ is defined as the cone of Griffiths non-negative polynomials.

Proposition 4.6 (Fulton-Lazarsfeld [Reference Fulton and Lazarsfeld11, Proposition A.3])

Let

$$ \begin{align*}P=\sum_{\lambda \in \Lambda(k, r)} a_\lambda(P) P_\lambda \quad\left(a_\lambda(P) \in \mathbb{Q}\right) \end{align*} $$

be a non-zero weighted homogeneous polynomial in $\mathbb {Q}\left[c_1, \ldots , c_r\right]$ . Then P lies in the Griffiths cone $\Pi (r)$ if and only if each of the Schur coefficients $a_\lambda (P)$ is non-negative.

In particular, for each $\lambda \in \Lambda (k,r)$ , one has

$$ \begin{align*} P_\lambda(B)=\sum_{\sigma, \tau \in S_k} \sum_{\rho \in[1, r]^k} p_{\rho \sigma \tau} B_{\rho_{\sigma(1)} \rho_{\tau(1)}} \cdots B_{\rho_{\sigma(k)} \rho_{\tau(k)}}, \end{align*} $$

where $p_{\rho \sigma \tau }=\sum _{1 \leq i, j \leq m} \left(\frac {1}{k!}\right)^2a_{ij}(\tau ) \overline {a_{ij}(\sigma )}$ with $(a_{ij}(\tau ))\in \mathrm {U}(m)$ , see [Reference Fulton and Lazarsfeld11, (A.6)]. Denote $T=[1,m]^2$ and $q_{\sigma t}:=\overline {a_t(\sigma )}$ for any $t\in T$ . Then

(4.13) $$ \begin{align} P_\lambda(B)=\left(\frac{1}{k!}\right)^2\sum_{\sigma, \tau \in S_k} \sum_{\rho \in[1, r]^k} \left(\sum_{t\in T}q_{\sigma t}\overline{q_{\tau t}}\right) B_{\rho_{\sigma(1)} \rho_{\tau(1)}} \cdots B_{\rho_{\sigma(k)} \rho_{\tau(k)}}. \end{align} $$

4.2.3. Weak positivity of Schur forms

We assume that $(E,h^E)$ is a strongly decomposably positive vector bundle of type I over a complex manifold X. By Theorem 4.3, for any $x\in X$ , there exists a decomposition $T_x^{1,0}X=U_x\oplus V_x$ such that the Chern curvature matrix R of $(E,h^E)$ has the form

(4.14) $$ \begin{align} R=-B\wedge\overline{B}^\top+A\wedge \overline{A}^\top \end{align} $$

with respect to a unitary frame, where B is a $r\times N$ matrix with $(0,1)$ -forms in $\overline {U^*_x}$ as entries and A is a $r\times N$ matrix with $(1,0)$ -forms in $V^*_x$ as entries. Moreover, $\mathrm {rank}(\mathbf {A})=r\cdot \dim V_x$ , $\mathrm {rank}(\mathbf {B})=r\cdot \dim U_x$ .

For each $\lambda \in \Lambda (k,r)$ , by (4.13), the Schur form $P_\lambda (c(E,h^E))$ is given by

$$ \begin{align*} P_\lambda(c(E,h^E))=\left(\frac{\sqrt{-1}}{2\pi}\right)^k\frac{1}{(k!)^2}\sum_{\sigma,\tau\in S_k}\sum_{\rho\in [1,r]^k}\left(\sum_{t\in T}q_{\sigma t}\overline{q_{\tau t}}\right)\cdot \bigwedge_{j=1}^k R_{\rho_{\sigma(j)}\overline{\rho_{\tau(j)}}}. \end{align*} $$

By (4.14), the Chern curvature matrix satisfies

$$ \begin{align*} R_{\rho_{\sigma(j)}\overline{\rho_{\tau(j)}}}&=(B_{\rho_{\tau(j)}c_j\bar{\beta_j}}\overline{B_{\rho_{\sigma(j)}c_j\overline{\alpha_j}}}+A_{\rho_{\tau(j)}c_j\alpha_j}\overline{A_{\rho_{\sigma(j)}c_j\beta_j}})dz^{\alpha_j}\wedge d\bar{z}^{\beta_j}\\ &=\sum_{c_j=1}^N(\overline{B_{\rho_{\sigma(j)}c_j}}\wedge B_{\rho_{\tau(j)}c_j}+A_{\rho_{\tau(j)}c_j}\wedge \overline{A_{\rho_{\sigma(j)}c_j}})\\ & =\sum_{c_j=1}^N\sum_{\epsilon_j\in\{0,1\}}(\overline{B_{\rho_{\sigma(j)}c_j}}\wedge B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j}\wedge \overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}, \end{align*} $$

which follows that

$$ \begin{align*} & \bigwedge_{j=1}^k R_{\rho_{\sigma(j)}\overline{\rho_{\tau(j)}}}=\bigwedge_{j=1}^k\sum_{c_j=1}^N\sum_{\epsilon_j\in\{0,1\}}(\overline{B_{\rho_{\sigma(j)}c_j}}\wedge B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j}\wedge \overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}\\& \quad =\sum_{c\in [1,N]^k}\sum_{\epsilon\in \{0,1\}^k}\bigwedge_{j=1}^k(\overline{B_{\rho_{\sigma(j)}c_j}}\wedge B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j}\wedge \overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}\\& \quad =\sum_{c\in [1,N]^k}\sum_{\epsilon\in \{0,1\}^k}\bigwedge_{j=1}^k(-1)^{\epsilon_j+1}(\overline{B_{\rho_{\sigma(j)}c_j}})^{\epsilon_j}\wedge (\overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}\wedge (B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j})^{1-\epsilon_j}\\& \quad =\sum_{c\in [1,N]^k}\sum_{\epsilon\in \{0,1\}^k}(-1)^{|\epsilon|+k}(-1)^{\frac{k(k-1)}{2}}\cdot\\&\qquad \left(\bigwedge_{j=1}^k(\overline{B_{\rho_{\sigma(j)}c_j}})^{\epsilon_j}\wedge (\overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}\right)\wedge \left(\bigwedge_{j=1}^k(B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j})^{1-\epsilon_j}\right), \end{align*} $$

where $|\epsilon |:=\sum _{j=1}^k \epsilon _j$ .

Recall that $\rho \in [1,r]^k$ , $t\in T$ , $c\in [1,N]^k$ and $\epsilon \in \{0,1\}^k$ . We obtain that

$$ \begin{align*} &P_\lambda(c(E,h^E))=\left(\frac{\sqrt{-1}}{2\pi}\right)^k\frac{1}{(k!)^2}(-1)^{\frac{k(k-1)}{2}}\sum_{\rho,t,c,\epsilon}(-1)^{|\epsilon|+k}\cdot\\& \quad \left(\sum_{\sigma\in S_k}q_{\sigma t}\bigwedge_{j=1}^k(\overline{B_{\rho_{\sigma(j)}c_j}})^{\epsilon_j}\wedge (\overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}\right)\wedge \left(\sum_{\tau\in S_k}\overline{q_{\tau t}}\bigwedge_{j=1}^k(B_{\rho_{\tau(j)}c_j})^{\epsilon_j}\wedge(A_{\rho_{\tau(j)}c_j})^{1-\epsilon_j}\right). \end{align*} $$

Now we set

(4.15) $$ \begin{align} \psi_{\rho t c\epsilon}:=\sum_{\sigma\in S_k}q_{\sigma t}\bigwedge_{j=1}^k(\overline{B_{\rho_{\sigma(j)}c_j}})^{\epsilon_j}\wedge (\overline{A_{\rho_{\sigma(j)}c_j}})^{1-\epsilon_j}, \end{align} $$

which is a $(|\epsilon |,k-|\epsilon |)$ -form. Hence,

(4.16) $$ \begin{align} P_\lambda(c(E,h^E))=\left(\frac{1}{2\pi}\right)^k\left(\frac{1}{k!}\right)^2(\sqrt{-1})^{k^2}\sum_{\rho,t,c,\epsilon}(-1)^{|\epsilon|+k}\psi_{\rho t c\epsilon}\wedge \overline{\psi_{\rho t c\epsilon}}. \end{align} $$

For any non-zero decomposable $(n-k,0)$ -form $\eta =\eta _1\wedge \cdots \wedge \eta _{n-k}$ , where $\eta _i,1\leq i\leq n-k$ , are $(1,0)$ -forms, we assume that $\eta _1,\cdots ,\eta _{i_0}\in U_x^*$ and $\eta _{i_0+1},\cdots ,\eta _{n-k}\in V_x^*$ . Now we can take local holomorphic coordinates $\{z^\alpha \}_{1\leq \alpha \leq n}$ around $x\in X$ such that

$$ \begin{align*} U_x^*= \mathrm{span}_{\mathbb{C}}\{dz^1|_x,\cdots, dz^{n_0}|_x\},\text{ with } dz^j|_x=\eta_j,\quad 1\leq j\leq i_0 \end{align*} $$

and

$$ \begin{align*} V_x^*= \mathrm{span}_{\mathbb{C}}\{dz^{n_0+1}|_x,\cdots, dz^{n}|_x\},\text{ with } dz^{n_0-i_0+j}|_x=\eta_j,\quad i_0+1\leq j\leq n-k, \end{align*} $$

and so $\psi _{\rho t c \epsilon }$ can be written in the following form:

$$ \begin{align*} \psi_{\rho t c \epsilon}=\sum_{1\leq\alpha_1<\cdots<\alpha_{|\epsilon|}\leq n_0\atop n_0+1\leq\beta_1<\cdots<\beta_{k-|\epsilon|}\leq n}\psi_{\alpha_1\cdots\alpha_{|\epsilon|}\bar{\beta}_1\cdots\bar{\beta}_{k-|\epsilon|}}dz^{\alpha_1}\wedge\cdots\wedge dz^{\alpha_{|\epsilon|}}&\\ &\qquad \wedge d\bar{z}^{\beta_1}\wedge\cdots\wedge d\bar{z}^{\beta_{k-|\epsilon|}}. \end{align*} $$

Then

(4.17) $$ \begin{align} & (\sqrt{-1})^{k^2}\sum_{\rho,t,c,\epsilon}(-1)^{|\epsilon|+k}\psi_{\rho t c\epsilon}\wedge \overline{\psi_{\rho t c\epsilon}}\wedge (\sqrt{-1})^{(n-k)^2}\eta\wedge\overline{\eta} \nonumber\\& \quad =(\sqrt{-1})^{k^2}\sum_{\rho,t,c,|\epsilon|=n_0-i_0}(-1)^{n_0-i_0+k}(\sqrt{-1})^{(n-k)^2}dz^1\wedge\cdots \wedge dz^{i_0}\wedge \nonumber\\& \qquad dz^{n_0+1}\wedge\cdots\wedge dz^{n-i_0+n-k}\wedge d\bar{z}^1\wedge \cdots\wedge d\bar{z}^{i_0}\wedge d\bar{z}^{n_0+1}\wedge\cdots\wedge d\bar{z}^{n-i_0+n-k}\wedge \nonumber\\& \qquad \left(\psi_{i_0+1, \cdots,{n_0},\overline{n_0-i_0+n-k+1},\cdots,\bar{n}}dz^{i_0+1}\wedge\cdots\wedge dz^{n_0}\wedge d\bar{z}^{n_0-i_0+n-k+1}\wedge\cdots\wedge d\bar{z}^{n}\right)\wedge\\& \qquad \left(\overline{\psi_{i_0+1, \cdots,{n_0},\overline{n_0-i_0+n-k+1},\cdots,\bar{n}}}d\bar{z}^{i_0+1}\wedge\cdots\wedge d\bar{z}^{n_0}\wedge d{z}^{n_0-i_0+n-k+1}\wedge\cdots\wedge d{z}^{n}\right) \nonumber\\& \quad =\sum_{\rho,t,c,|\epsilon|=n_0-i_0}|\psi_{i_0+1, \cdots,{n_0},\overline{n_0-i_0+n-k+1},\cdots,\bar{n}}|^2\cdot \nonumber\\&\qquad (\sqrt{-1})^{n^2}dz^1\wedge \cdots\wedge dz^n\wedge d\bar{z}^1\wedge\cdots\wedge d\bar{z}^n,\nonumber \end{align} $$

which is a non-negative volume form. By (4.16), the Schur form $P_\lambda (c(E,h^E))$ is weakly non-negative.

We show the weak positivity of Schur form $P_\lambda (c(E,h^E))$ using a proof by contradiction. Specifically, we derive a contradiction when assuming $P_\lambda (c(E,h^E))\wedge (\sqrt {-1})^{(n-k)^2}\eta \wedge \bar {\eta }=0$ .

By (4.17), one knows that

$$ \begin{align*}P_\lambda(c(E,h^E))\wedge(\sqrt{-1})^{(n-k)^2}\eta\wedge\bar{\eta}=0\end{align*} $$

if and only if

(4.18) $$ \begin{align} \psi_{i_0+1, \cdots,{n_0},\overline{n_0-i_0+n-k+1},\cdots,\bar{n}}=0 \end{align} $$

for any $\rho ,t,c,|\epsilon |=n_0-i_0$ .

Now we take a special vector $\epsilon =(\underbrace {1,\cdots ,1}_{n_0-i_0},0\cdots ,0)$ and denote $j_0=n_0-i_0$ , by (4.15). Then

$$ \begin{align*} \psi_{\rho t c\epsilon}&=\sum_{\sigma \in S_k} q_{\sigma t}\overline{B_{\rho_{\sigma(1)}c_1}}\wedge \cdots\wedge \overline{B_{\rho_{\sigma(j_0)}c_{j_0}}}\wedge \overline{A_{\rho_{\sigma(j_0+1)}c_{j_0+1}}}\wedge\cdots \wedge \overline{A_{\rho_{\sigma(k)}c_k}}. \end{align*} $$

Combining with (4.18), one has

(4.19) $$ \begin{align} 0 &=\psi_{i_0+1, \cdots,{n_0},\overline{n_0-i_0+n-k+1},\cdots,\bar{n}} \nonumber\\ &=\sum_{\tau_1\in S_{j_0}}\sum_{\tau_2\in S_{k-j_0}}\sum_{\sigma \in S_k} \mathrm{sgn}(\tau_1)\mathrm{sgn}(\tau_2)q_{\sigma t}\cdot \\ &\overline{B_{\rho_{\sigma(1)}c_1\overline{\tau_1(i_0+1)}}}\cdots\overline{B_{\rho_{\sigma(j_0)}c_{j_0}\overline{\tau_1(n_0)}}}\cdot \overline{A_{\rho_{\sigma(j_0+1)}c_{j_0+1}{\tau_2(j_0+n-k+1)}}}\cdots \overline{A_{\rho_{\sigma(k)}c_k{\tau_2(n)}}}.\nonumber \end{align} $$

Since $\mathrm {rank}(\mathbf {A})=r\cdot \dim V_x$ , $\mathrm {rank}(\mathbf {B})=r\cdot \dim U_x$ , without loss of generality, we assume that the submatrices

$$ \begin{align*}\mathbf{B}'=(\mathbf{B}_{c,i\alpha})_{1\leq c\leq r\dim U_x, 1\leq i\leq r,1\leq \alpha\leq \dim U_x}\end{align*} $$

and

$$ \begin{align*} \mathbf{A}'=(\mathbf{A}_{c,j\alpha})_{1\leq c\leq r\dim V_x, 1\leq j\leq r,1\leq \alpha\leq \dim V_x} \end{align*} $$

of $\mathbf {B}$ and $\mathbf {A}$ are inverse. By (4.19) and note that $\mathbf {B}^{\prime }_{c,i\alpha }=B_{ic\bar {\alpha }}$ and $\mathbf {A}^{\prime }_{c,j\alpha }=A_{jc\alpha }$ , one has

(4.20) $$ \begin{align} 0&=\sum_{\tau_1\in S_{j_0}}\sum_{\tau_2\in S_{k-j_0}}\sum_{c_1,\cdots,c_{j_0}=1}^{r n_0}\sum_{c_{j_0+1},\cdots,c_{k}=1}^{r (n-n_0)}\sum_{\sigma \in S_k} \mathrm{sgn}(\tau_1)\mathrm{sgn}(\tau_2)q_{\sigma t}\cdot \nonumber\\ &\overline{\mathbf{B}^{\prime}_{c_1,\rho_{\sigma(1)}{\tau_1(i_0+1)}}}\cdots\overline{\mathbf{B}^{\prime}_{c_{j_0},\rho_{\sigma(j_0)}{\tau_1(n_0)}}}\cdot \overline{\mathbf{A}^{\prime}_{c_{j_0+1},\rho_{\sigma(j_0+1)}{\tau_2(j_0+n-k+1)}}}\cdots \overline{\mathbf{A}^{\prime}_{c_k,\rho_{\sigma(k)}{\tau_2(n)}}}\cdot \nonumber\\ &\overline{\mathbf{A}{^{\prime}-1}_{l_k\beta_k,c_k}}\cdots \overline{\mathbf{A}{^{\prime}-1}_{l_{j_0+1}\beta_{j_0+1},c_{j_0+1}}}\overline{\mathbf{B}{^{\prime}-1}_{l_{j_0}\beta_{j_0},c_{j_0}}}\cdots \overline{\mathbf{B}{^{\prime}-1}_{l_{1}\beta_{1},c_{1}}} \\ &=\sum_{\tau_1\in S_{j_0}}\sum_{\tau_2\in S_{k-j_0}}\sum_{\sigma \in S_k} \mathrm{sgn}(\tau_1)\mathrm{sgn}(\tau_2)q_{\sigma t}\cdot\delta_{\rho_{\sigma(k)}l_k}\delta_{\tau_2(n)\beta_k}\cdots\delta_{\rho_{\sigma(j_0+1)}l_{j_0+1}}\cdot \nonumber\\ &\delta_{\tau_2(j_0+n-k+1)\beta_{j_0+1}}\delta_{\rho_{\sigma(j_0)}l_{j_0}}\delta_{\tau_1(n_0)\beta_{j_0}}\cdots\delta_{\rho_{\sigma(1)}l_1}\delta_{\tau_1(i_0+1)\beta_1}\nonumber \end{align} $$

for any $(\beta _1,\cdots ,\beta _{j_0})\in [1,n_0]^{j_0}$ , $(\beta _{j_0+1},\cdots ,\beta _k)\in [n_0+1,n]^{n-j_0}$ and $(l_1,\cdots ,l_k)\in [1,r]^k$ .

By taking

$$ \begin{align*}\beta_s= \begin{cases} i_0+s& 1\leq s\leq j_0 ,\\ n-k+s& j_0+1\leq s\leq k, \end{cases} \end{align*} $$

(4.20) becomes

(4.21) $$ \begin{align} \sum_{\sigma\in S_k} q_{\sigma t}\delta_{\rho_{\sigma(1)}l_1}\cdots\delta_{\rho_{\sigma(k)}l_k}=0 \end{align} $$

for any $\rho ,l\in [1,r]^k$ and $t\in T$ .

Remark 4.7. Note that (4.21) holds if and only if $\psi _{\rho t c \epsilon }=0$ for any $\rho ,t,c,\epsilon $ . In fact, if $\psi _{\rho t c \epsilon }=0$ , then (4.18) holds and follows (4.21). Conversely, if (4.21) holds, then

$$ \begin{align*} \psi_{\rho t c \epsilon}=\sum_{l\in[1,r]^k} (\sum_{\sigma\in S_k} q_{\sigma t}\delta_{\rho_{\sigma(1)}l_1}\cdots\delta_{\rho_{\sigma(k)}l_k})\bigwedge_{j=1}^k\overline{B_{l_jc_j}}^{\epsilon_j}\wedge \overline{A_{l_jc_j}}^{1-\epsilon_j}=0. \end{align*} $$

For $k\leq r$ , one can take $\rho _i=l_i=i$ for $1\leq i\leq k$ . Thus,

$$ \begin{align*} 0=\sum_{\sigma\in S_k} q_{\sigma t}\delta_{\rho_{\sigma(1)}l_1}\cdots\delta_{\rho_{\sigma(k)}l_k}=q_{\mathrm{Id},t} \end{align*} $$

for any $t\in T$ , which is a contradiction since $(q_{\mathrm {Id},t})_{t\in T}\in \mathrm {U}(m)$ is a unitary matrix. Hence, all $(k,k)$ -Schur forms $P_\lambda (c(E,h^E))$ are weakly positive for any $k\leq r$ . In particular, all Chern forms $c_i(E,h^E),1\leq i\leq r$ , are weakly positive.

For general k and r, we take $l_1,\cdots ,l_k$ in (4.21) to be

$$ \begin{align*} l_i=\rho_i, \text{ for }1\leq i\leq k, \end{align*} $$

and (4.21) implies that

(4.22) $$ \begin{align} \sum_{\rho\in [1,r]^k}\sum_{\sigma\in S_k}\chi_\lambda (\sigma)\delta_{\rho_1\rho_{\sigma(1)}}\cdots\delta_{\rho_k\rho_{\sigma(k)}}=0, \end{align} $$

where

$$ \begin{align*}\chi_\lambda(\sigma)=\mathrm{Tr}(\overline{q_{\sigma t}})=\sum_{i=1}^m a_{ii}(\sigma)\end{align*} $$

is the character of the representation $\phi _\lambda (\sigma )=(a_{ij}(\sigma ))\in \mathrm {U}(m)$ corresponding to the partition $\lambda $ . From [Reference Fulton and Lazarsfeld11, (A.5)], (4.22) is equivalent to

(4.23) $$ \begin{align} P_\lambda (I_r)=0. \end{align} $$

Here, $P_\lambda (\bullet )$ denotes the invariant polynomial corresponding to the Schur function $P_\lambda $ under the isomorphism $\mathrm {I}(r)\cong \mathbb {Q}(c_1,\cdots ,c_r)$ .

Denote by $x_1,\cdots ,x_r$ the Chern roots, which are defined by

$$ \begin{align*} \sum_{j=0}^r c_jt^j=(1+tx_1)(1+tx_2)\cdots(1+tx_r). \end{align*} $$

Recall that the Schur polynomial is defined by

$$ \begin{align*} P_\lambda\left(c_1, \ldots, c_r\right)&=\operatorname{det}\left(c_{\lambda_j-j+l}\right)_{1 \leqslant j, l \leqslant k}, \end{align*} $$

where $\lambda =(\lambda _1,\cdots ,\lambda _k)\in \Lambda (k,r)$ is a partition satisfying

$$ \begin{align*}\sum_{i=1}^k\lambda_i=k \text{ and } r\geq \lambda_1\geq\cdots\geq\lambda_k\geq 0.\end{align*} $$

Denote by $\lambda '$ the conjugate partition to the partition $\lambda $ (see, for example, [Reference Fulton and Harris12, Section 4.1, Page 45]). Then

(4.24) $$ \begin{align} \lambda'=(\lambda_1',\cdots,\lambda_r'),\quad\text{ with } \sum_{i=1}^r\lambda^{\prime}_i=k\text{ and } \lambda^{\prime}_1\geq\cdots\geq\lambda^{\prime}_r\geq 0. \end{align} $$

The second Jacobi-Trudi identity (or Giambell’s formula) gives

(4.25) $$ \begin{align} P_\lambda\left(c_1, \ldots, c_r\right)&=\operatorname{det}\left(c_{\lambda_j-j+l}\right)_{1 \leqslant j, l \leqslant k}=s_{\lambda'}(x_1,\cdots,x_r); \end{align} $$

see, for example, [Reference Fulton and Harris12, Page 455, (A.6)], where

(4.26) $$ \begin{align} s_{\lambda'}(x_1,\cdots,x_r):=\frac{\left|\begin{array}{cccc}x_1^{\lambda^{\prime}_1+r-1} & x_2^{\lambda^{\prime}_1+r-1} & \ldots & x_r^{\lambda^{\prime}_1+r-1} \\ x_1^{\lambda^{\prime}_2+r-2} & x_2^{\lambda^{\prime}_2+r-2} & \ldots & x_r^{\lambda^{\prime}_2+r-2} \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{\lambda^{\prime}_r} & x_2^{\lambda^{\prime}_r} & \ldots & x_r^{\lambda^{\prime}_r}\end{array}\right|}{\prod_{1 \leq i<j \leq r}\left(x_i-x_j\right)}. \end{align} $$

In particular, we have

(4.27) $$ \begin{align} P_\lambda(I_r)&=P_\lambda\left(c_1=C^1_r,\cdots, c_i=C_r^i,\cdots,c_r=C^r_r\right) \nonumber\\ &=s_{\lambda'}(1,\cdots,1)\\ &=\prod_{1 \leq i<j \leq r} \frac{\lambda^{\prime}_i-\lambda^{\prime}_j+j-i}{j-i}\geq 1, \nonumber \end{align} $$

where the third equality follows from [Reference Fulton and Harris12, Page 461, (ii)]. Hence, $ P_\lambda (I_r)\neq 0$ , which contradicts (4.23). Thus,

$$ \begin{align*}P_\lambda(c(E,h^E))\wedge(\sqrt{-1})^{(n-k)^2}\eta\wedge\bar{\eta}>0\end{align*} $$

for any non-zero decomposable $(n-k,0)$ -form $\eta $ , and so $P_\lambda (c(E,h^E))$ is a weakly positive $(k,k)$ -form.

We obtain the following.

Theorem 4.8. Let $(E,h^E)$ be a strongly decomposably positive vector bundle of type I over a complex manifold X, $\mathrm {rank}E=r$ , and $\dim X=n$ . Then the Schur form $P_\lambda (c(E,h^E))$ is weakly positive for any partition $\lambda \in \Lambda (k,r)$ , $k\leq n$ and $k\in \mathbb {N}$ .

In particular, if $(E,h^E)$ is Nakano positive, then the Chern curvature matrix has the form

$$ \begin{align*} R=-B\wedge\overline{B}^\top. \end{align*} $$

By considering $A=0$ in (4.15), then

$$ \begin{align*} \psi_{\rho t c \epsilon_1}=\sum_{\sigma\in S_k}q_{\sigma t}\bigwedge_{j=1}^k\overline{B_{\rho_{\sigma(j)}c_j}},\quad \epsilon_1=(1,\cdots,1) \end{align*} $$

and

$$ \begin{align*} \psi_{\rho t c \epsilon}=0,\text{ for any }\epsilon\neq \epsilon_1. \end{align*} $$

By (4.16), one has

(4.28) $$ \begin{align} P_\lambda(c(E,h^E))=\left(\frac{1}{2\pi}\right)^k\left(\frac{1}{k!}\right)^2(\sqrt{-1})^{k^2}\sum_{\rho,t,c}\psi_{\rho t c \epsilon_1}\wedge \overline{\psi_{\rho t c\epsilon_1}}, \end{align} $$

where $\psi _{\rho t c\epsilon _1}$ is a $(k,0)$ -form. For any non-zero $(n-k,0)$ -form $\eta $ , one has

$$ \begin{align*} & \quad P_\lambda(c(E,h^E))\wedge (\sqrt{-1})^{(n-k)^2}\eta\wedge\overline{\eta}\\ &=\left(\frac{1}{2\pi}\right)^k\left(\frac{1}{k!}\right)^2(\sqrt{-1})^{n^2}\sum_{\rho,t,c}\psi_{\rho t c \epsilon_1}\wedge\eta\wedge \overline{\psi_{\rho t c\epsilon_1}\wedge \eta}, \end{align*} $$

which is a non-negative volume form, and so $ P_\lambda (c(E,h^E))$ is non-negative. Moreover,

(4.29) $$ \begin{align} P_\lambda(c(E,h^E))\wedge (\sqrt{-1})^{(n-k)^2}\eta\wedge\overline{\eta}=0 \end{align} $$

if and only if

(4.30) $$ \begin{align} \psi_{\rho t c \epsilon_1}\wedge \eta=0 \end{align} $$

for any $\rho \in [1,r]^k$ , $t\in T$ , $c\in [1,N]^k$ . By the expression of $\psi _{\rho t c\epsilon _1}$ , (4.30) becomes

(4.31) $$ \begin{align} 0&=\psi_{\rho t c \epsilon_1}\wedge \eta \nonumber\\ &=\sum_{\sigma\in S_k} q_{\sigma t}\overline{B_{\rho_{\sigma(1)}c_1}}\wedge \cdots\wedge \overline{B_{\rho_{\sigma(k)}c_k}}\wedge \eta\\ &=\sum_{\sigma\in S_k} q_{\sigma t}\overline{B_{\rho_{\sigma(1)}c_1\bar{\alpha}_1}} \cdots \overline{B_{\rho_{\sigma(k)}c_k\bar{\alpha}_k}} dz^{\alpha_1}\wedge\cdots\wedge dz^{\alpha_k}\wedge \eta.\nonumber \end{align} $$

Since $(E,h^E)$ is Nakano positive, by Corollary 4.4, we can take B such that $\mathbf {B}$ is invertible. Multiplying (4.31) by $(\mathbf {B}^{-1})_{c_1,l_1\beta _1}\cdots (\mathbf {B}^{-1})_{c_k,l_k\beta _k}$ and summing on $c_1,\cdots ,c_k$ , one has

$$ \begin{align*} \left( \sum_{\sigma\in S_k} q_{\sigma t}\delta_{\rho_{\sigma(1)}l_1}\cdots\delta_{\rho_{\sigma(k)}l_k}\right)dz^{\beta_1}\wedge \cdots\wedge dz^{\beta_k}\wedge \eta=0 \end{align*} $$

for any $l=(l_1,\cdots ,l_k)\in [1,r]^k$ and $\beta =(\beta _1,\cdots ,\beta _k)\in [1,n]^k$ . By choosing $\beta _1,\cdots ,\beta _k$ such that $dz^{\beta _1}\wedge \cdots \wedge dz^{\beta _k}\wedge \eta \neq 0$ ,

(4.32) $$ \begin{align} \sum_{\sigma\in S_k} q_{\sigma t}\delta_{\rho_{\sigma(1)}l_1}\cdots\delta_{\rho_{\sigma(k)}l_k}=0, \end{align} $$

which is exactly (4.21). By Remark 4.7, (4.32) is equivalent to $\psi _{\rho t c\epsilon _1}=0$ . Hence, (4.29) is equivalent to (4.32), which follows that $P_\lambda (I_r)=0$ ; see (4.23). By (4.27), $P_\lambda (I_r)\neq 0$ , so we get a contradiction. Thus, $P_\lambda (c(E,h^E))$ is a positive $(k,k)$ -form. Similarly, if $(E,h^E)$ is dual Nakano positive, then $P_\lambda (c(E,h^E))$ is also a positive $(k,k)$ -form.

Hence, we can give an algebraic proof of the following positivity of Schur forms for (dual) Nakano positive vector bundles.

Theorem 4.9 (Finski [Reference Finski9, Theorem 1.1])

Let $(E, h^E)$ be a (dual) Nakano positive (respectively non-negative) vector bundle of rank r over a complex manifold X of dimension n. Then for any $k \in \mathbb {N}$ , $k \leqslant n$ , and $\lambda \in \Lambda (k, r)$ , the $(k, k)$ -form $P_\lambda \left(c\left(E, h^E\right)\right)$ is positive (respectively non-negative).

Remark 4.10. Note that S. Finski proved the above positivity of Schur forms by using the following two steps: the first one is a refinement of the determinantal formula of Kempf-Laksov on the level of differential forms, which expresses Schur forms as a certain pushforward of the top Chern form of a Hermitian vector bundle obtained as a quotient of the tensor power of $(E,h^E)$ , and the second one is to show the positivity of the top Chern form of a (dual) Nakano positive vector bundle. Our method here is an algebraic proof by analyzing the vanishing of Schur forms, which is very different from S. Finski’s approach.

5.1. A criterion of type II positivity

Let $(E,h^E)$ be a strongly decomposably positive vector bundle of type II; see Definition 2.7. By Corollary 4.4, with respect to a unitary frame $\{e_1,\cdots ,e_{r_1}\}$ of $(E_{1,x},h^E|_{E_{1,x}})$ , and a unitary frame $\{e_{r_1+1},\cdots ,e_r\}$ of $(E_{2,x},h^E|_{E_{2,x}})$ at $x\in X$ , one has

$$ \begin{align*} R^{E}_x|_{E_{1,x}}=-B_1\wedge\overline{B_1}^\top, \quad R^{E}_x|_{E_{2,x}}=A_2\wedge \overline{A_2}^\top, \end{align*} $$

with $\mathrm {rank}(\mathbf {B}_1)=\dim (E_{1,x})\cdot n$ and $\mathrm {rank}(\mathbf {A}_2)=\dim (E_{2,x})\cdot n$ , where $B_1=((B_1)_{ip})_{1\leq i\leq r_1,1\leq j\leq N_1}$ is a matrix with $(0,1)$ -forms as entries and $A_2=((A_2)_{ip})_{r_1+1\leq i\leq r,1\leq j\leq N_2}$ is a matrix with $(1,0)$ -forms as entries. The matrices $\mathbf {B}_1$ and $\mathbf {A}_2$ are defined in (4.10) and (4.11), respectively. Now we define the matrices $A_{r\times N}$ and $B_{r\times N}$ ( $N=\max \{N_1,N_2\}$ ) by

(5.1) $$ \begin{align} B=\begin{pmatrix} B_1& 0\\ 0& 0 \end{pmatrix},\quad A=\begin{pmatrix} 0&0 \\ A_2&0 \end{pmatrix}. \end{align} $$

Then

$$ \begin{align*} \mathrm{rank}(\mathbf{B})=\mathrm{rank}(\mathbf{B}_1)=r_1\cdot n,\quad \mathrm{rank}(\mathbf{A})=\mathrm{rank}(\mathbf{A}_2)=(n-r_1)\cdot n \end{align*} $$

and

$$ \begin{align*} R^E_x=\begin{pmatrix} -B_1\wedge \overline{B_1}^\top&0 \\ 0& A_2\wedge\overline{A_2}^\top \end{pmatrix}=-B\wedge\overline{B}^\top+A\wedge \overline{A}^\top. \end{align*} $$

For the matrix $\mathbf {B}=(B_{ip\bar {\alpha }})_{i\alpha ,p}$ , we can associate it with another matrix $\mathcal {B}$ by

$$ \begin{align*} \mathcal{B}=(\mathcal{B}_{\alpha p})=\left(\sum_{i=1}^rB_{ip\bar{\alpha}}e_i\right)_{\alpha p}, \end{align*} $$

which is a $n\times N$ matrix with elements of E as entries. Similarly, we can define a $n\times N$ matrix by

$$ \begin{align*} \mathcal{A}=(\mathcal{A}_{\alpha p})=\left(\sum_{i=1}^r A_{ip\alpha}e_i\right)_{\alpha p}. \end{align*} $$

We define

$$ \begin{align*} \{\mathcal{B}\}:=\left\{\sum_{i=1}^rB_{ip\bar{\alpha}}e_i,1\leq p\leq N, 1\leq\alpha\leq n\right\} \end{align*} $$

and

$$ \begin{align*} \{\mathcal{A}\}:=\left\{\sum_{i=1}^rA_{ip{\alpha}}e_i,1\leq p\leq N, 1\leq\alpha\leq n\right\}. \end{align*} $$

By the definitions of the matrices A and B, one has

$$ \begin{align*} \mathrm{span}_{\mathbb{C}}\{\mathcal{A}\}\perp \mathrm{span}_{\mathbb{C}}\{\mathcal{B}\}. \end{align*} $$

Hence, if $(E,h^E)$ is a strongly decomposably positive vector bundle of type II, then there are two $r\times N$ -matrices A, B of $(1,0)$ -forms and $(0,1)$ -forms, respectively, such that with respect to a unitary frame $\{e_i\}_{1\leq i\leq r}$ of $E_x$ ,

(5.2) $$ \begin{align} R^E_x=-B\wedge \overline{B}^\top+A\wedge\overline{A}^\top, \end{align} $$

and

(5.3) $$ \begin{align} \mathrm{span}_{\mathbb{C}}\{\mathcal{A}\}\perp \mathrm{span}_{\mathbb{C}}\{\mathcal{B}\}. \end{align} $$

Moreover, the ranks of $\mathbf {A}$ and $\mathbf {B}$ satisfy

(5.4) $$ \begin{align} \mathrm{rank}(\mathbf{B})=r_1\cdot n, \quad \mathrm{rank}(\mathbf{A})=(r-r_1)\cdot n. \end{align} $$

Remark 5.1. If we consider a new unitary frame $\widetilde {e}=e\cdot a$ for a unitary matrix $a\in \mathrm {U}(r)$ , by Remark 4.1, one has

(5.5) $$ \begin{align} \widetilde{R^E_x}=-\widetilde{B}\wedge \overline{\widetilde{B}}^\top+\widetilde{A}\wedge\overline{\widetilde{A}}^\top, \end{align} $$

with $\widetilde {B}=a^{-1}\cdot B$ and $\widetilde {A}=a^{-1}\cdot A$ . Moreover, one has

$$ \begin{align*} \widetilde{\mathcal{B}}_{\alpha p}=\sum_{i=1}^r\widetilde{B}_{ip\bar{\alpha}}\widetilde{e}_i=\sum_{i,j=1}^r(a^{-1})_{ij}B_{jp\bar{\alpha}}\widetilde{e}_i=\sum_{j=1}^r B_{jp\bar{\alpha}}{e}_j=\mathcal{B}_{\alpha p}. \end{align*} $$

Similarly, $\widetilde {\mathcal {A}}_{\alpha p}=\mathcal {A}_{\alpha p}$ . Hence, $\mathcal {A}$ and $\mathcal {B}$ are independent of the unitary frame. One can also check that

$$ \begin{align*} \mathrm{rank}(\widetilde{\mathbf{B}})=\mathrm{rank}(\mathbf{B})=r_1\cdot n,\quad \mathrm{rank}(\widetilde{\mathbf{A}})=\mathrm{rank}(\mathbf{A})=(r-r_1)\cdot n. \end{align*} $$

In a word, we show that (5.2)–(5.4) hold for any unitary frame.

Conversely, we assume that (5.2)–(5.4) hold for some unitary frame of $E_x$ , $x\in X$ . Set

$$ \begin{align*} E_{1,x}:=\mathrm{span}_{\mathbb{C}}\{\mathcal{B}\},\quad E_{2,x}:=\mathrm{span}_{\mathbb{C}}\{\mathcal{A}\}. \end{align*} $$

Let $\{e_1,\cdots , e_{r^{\prime }_1}\}$ be a unitary frame of $E_{1,x}$ and $\{e_{r_1'+1},\cdots , e_{r'}\}$ be a unitary frame of $E_{2,x}$ . Since $E_{1,x}\perp E_{2,x}$ , then $\{e_i\}_{1\leq i\leq r'}$ is a unitary frame of $E_{1,x}\oplus E_{2,x}$ . Now we can extend the frame $\{e_i\}_{1\leq i\leq r'}$ and get a unitary frame $\{e_i\}_{1\leq i\leq r}$ of $E_x$ . By Remark 5.1, (5.2)–(5.4) also hold for this unitary frame $\{e_i\}_{1\leq i\leq r}$ . Hence,

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}} e_j\otimes \overline{e_i}=\sum_{p=1}^N(-\mathcal{B}_{\beta p}\otimes \overline{\mathcal{B}_{\alpha p}}+\mathcal{A}_{\alpha p}\otimes \overline{\mathcal{A}_{\beta p}}). \end{align*} $$

So

$$ \begin{align*} R^E_x|_{E_{1,x}}=-B\wedge\overline{B}^\top,\,\, R^E_x|_{E_{2,x}}=A\wedge \overline{A}^\top \end{align*} $$

and

$$ \begin{align*} R_{i\bar{j}\alpha\bar{\beta}}=0\text{ for any } (i,j) \text{ or } (j,i)\in [1,r_1']\times [r_1'+1,r]. \end{align*} $$

By (5.4), one has

$$ \begin{align*} r_1'\geq r_1,\quad r-r_1'\geq r'-r^{\prime}_1\geq r-r_1, \end{align*} $$

which follows that

$$ \begin{align*} r_1=r_1',\quad r'=r. \end{align*} $$

Hence, $E_x=E_{1,x}\oplus E_{2,x}$ . By Corollary 4.4, we obtain that $R_{i\bar {j}\alpha \bar {\beta }}u^{i\alpha }\overline {u^{j\beta }}>0$ for any non-zero $u=u^{i\alpha } e_i\otimes \partial _\alpha \in E_{1,x}\otimes T^{1,0}_xX$ , $R_{i\bar {j}\alpha \bar {\beta }}v^{i\bar {\beta }}\overline {v^{j\bar {\alpha }}}>0$ for any non-zero $v=v^{i\bar {\beta }}e_i\otimes \partial _{\bar {\beta }}\in E_{2,x}\otimes T^{0,1}_xX$ . Thus, $(E,h^E)$ is a strongly decomposably positive vector bundle of type II.

In a word, we obtain a criterion of a strongly decomposably positive vector bundle of type II.

5.2. Positivity of Schur forms

In this subsection, we consider the positivity of Schur forms for strongly decomposably positive vector bundles of type II.

Let E and F be two holomorphic vector bundles over a complex manifold X, $\mathrm {rank}(E)=r$ and $\mathrm {rank}(F)=q$ . Let $x_1,\cdots ,x_{r}$ denote the Chern roots of E. For any partition $\lambda '$ satisfying (4.24), we denote

$$ \begin{align*} s_{\lambda'}(c(E)):=s_{\lambda'}(x_1,\cdots,x_r)\in \mathrm{H}^{2k}(X,\mathbb{R}), \end{align*} $$

where $s_{\lambda '}(x_1,\cdots ,x_r)$ is defined in (4.26), which is also called a Schur class. Similarly, one can define the cohomology classes $s_{\lambda '}(c(F))$ and $s_{\lambda '}(c(E\oplus F))$ . For these cohomology classes, by Littlewood-Richardson rule (see [Reference Billey, Rhoades and Tewari1, Proposition 3.3 (3.14)]), one has

$$ \begin{align*} s_{\lambda'}(c(E\oplus F))=\sum_{\mu',\nu'}c^{\lambda'}_{\mu'\nu'}s_{\mu'}(c(E))s_{\nu'}(c(F)), \end{align*} $$

where $c^{\lambda '}_{\mu '\nu '}$ is a Littlewood-Richardson coefficient. One can refer to [Reference Fulton10, Chapter 5] for more details on the Littlewood-Richardson coefficients. By (4.25), the Schur class $P_\lambda (c(E\oplus F))$ of the direct sum $E\oplus F$ satisfies

(5.6) $$ \begin{align} P_\lambda(c(E\oplus F))=\sum_{\mu',\nu'}c^{\lambda'}_{\mu'\nu'}P_\mu(c(E))P_\nu(c(F))=\sum_{\mu,\nu}c^\lambda_{\mu\nu}P_\mu(c(E))P_\nu(c(F)), \end{align} $$

where $\lambda ,\mu ,\nu $ are the conjugate partitions to $\lambda ',\mu ',\nu '$ , respectively. The last equality follows from the conjugation symmetry $c^{\lambda '}_{\mu '\nu '}=c^\lambda _{\mu \nu }$ ; see, for example, [Reference Stembridge19, Page 115].

Let $h^E$ and $h^F$ be Hermitian metrics on E and F, respectively. The direct sum $E\oplus F$ is equipped with the natural metric $h^E\oplus h^F$ . Now we can prove (5.6) on the level of differential forms.

Proposition 5.3. For any $\lambda \in \Lambda (k,r)$ , one has

$$ \begin{align*} P_\lambda(c(E\oplus F,h^E\oplus h^F))=\sum_{\mu,\nu}c^\lambda_{\mu\nu}P_\mu(c(E,h^{E}))\wedge P_\nu(c(F,h^{F})). \end{align*} $$

Proof. We follow the method in the proofs of [Reference Guler13, Proposition 3.1] and [Reference Diverio and Fagioli6, Theorem 3.5]. From the definition of total Chern form, one has

$$ \begin{align*} c(E\oplus F,h^E\oplus h^F)=c(E,h^E)\wedge c(F,h^F). \end{align*} $$

Recall that $P_\lambda \left(c_1, \ldots , c_r\right)=\operatorname {det}\left(c_{\lambda _i-i+j}\right)_{1 \leqslant i, j \leqslant k}$ , so that

$$ \begin{align*} & P_\lambda(c(E\oplus F,h^E\oplus h^F))- \sum_{\mu,\nu}c^\lambda_{\mu\nu}P_\mu(c(E,h^{E}))\wedge P_\nu(c(F,h^{F}))\\& \quad =\sum_{i_1+2i_2+\cdots+ri_r\atop+j_1+2j_2+\cdots+qj_q=k}f_{i_1\cdots i_r j_1\cdots j_{q}}c_1(E,h^E)^{i_1}\wedge \cdots\wedge c_r(E,h^E)^{i_r}\\ &\wedge c_1(F,h^F)^{j_1}\wedge \cdots\wedge c_q(F,h^F)^{j_q}, \end{align*} $$

where the universal coefficients $f_{i_1\cdots i_rj_1\dots j_q}$ do not depend on $E,F$ and X, but just depend $r,q$ , $P_\lambda $ . By (5.6), then the cohomology class satisfies

$$ \begin{align*} &\left[\sum_{i_1+2i_2+\cdots+ri_r\atop+j_1+2j_2+\cdots+qj_q=k}f_{i_1\cdots i_r j_1\cdots j_{q}}c_1(E,h^E)^{i_1}\wedge \cdots\wedge c_r(E,h^E)^{i_r} \wedge c_1(F,h^F)^{j_1}\wedge \cdots\wedge c_q(F,h^F)^{j_q}\right]=0. \end{align*} $$

Now we can take X as any n-dimensional projective manifold and fix an ample line bundle A on X. Let $\omega _A$ be a metric on A with positive curvature. For $m_1,\cdots ,m_r,m_{r+1},\cdots ,m_{r+q}$ positive integers, we define

$$ \begin{align*} E=A^{\otimes m_1}\oplus \cdots\oplus A^{\otimes m_r},\quad F=A^{\otimes m_{r+1}}\oplus\cdots\oplus A^{\otimes m_{r+q}}. \end{align*} $$

By the same proof as in [Reference Diverio and Fagioli6, Page 14], one can show all universal coefficients $f_{i_1\cdots i_r j_1\cdots j_{q}}$ vanish, which follows that

$$ \begin{align*} P_\lambda(c(E\oplus F,h^E\oplus h^F))- \sum_{\mu,\nu}c^\lambda_{\mu\nu}P_\mu(c(E,h^{E}))\wedge P_\nu(c(F,h^{F}))=0, \end{align*} $$

which completes the proof.

Now we assume $(E,h^E)$ is a strongly decomposably positive vector bundle of type II. For any $x\in X$ , there exists an orthogonal decomposition of $E_x$ ,

$$ \begin{align*} E_x=E_{1,x}\oplus E_{2,x}, \end{align*} $$

and the Chern curvature $R^E_x$ has the form

$$ \begin{align*} R^E_x=\begin{pmatrix} R^{E}_x|_{E_{1,x}}&0 \\ 0& R^{E}_x|_{E_{2,x}} \end{pmatrix}. \end{align*} $$

Let $\{e_1,\cdots , e_{r_1}\}$ be a unitary frame of $(E_{1,x},h^E|_{E_{1,x}})$ and $\{e_{r_1+1},\cdots , e_{r}\}$ be a unitary frame of $(E_{2,x},h^E|_{E_{2,x}})$ . Let $(U,\{z^\alpha \}_{1\leq \alpha \leq n})$ be a local coordinate neighborhood around x and denote by $E_1=U\times E_{1,x}$ the locally trivial bundle; $\{e_i\}_{1\leq i\leq r_1}$ also gives a frame of $E_1$ . Now we define the following Hermitian metric on $E_1$ by

$$ \begin{align*} h^{E_1}( e_i,e_j):= \delta_{ij}- R_{i\bar{j}\alpha\bar{\beta}} z^\alpha\bar{z}^\beta,\quad 1\leq i,j\leq r_1, \end{align*} $$

which is a Hermitian metric by taking U small enough. Then $(E_1,h^{E_1})$ is a Hermitian vector bundle around x and satisfies

$$ \begin{align*}R^{E_1}_x=R^E_x|_{E_{1,x}}. \end{align*} $$

Similarly, one can define a Hermitian vector bundle $(E_2,h^{E_2})$ such that $ R^{E_2}_x=R^E_x|_{E_{2,x}}. $ Hence,

$$ \begin{align*} c(E_1\oplus E_2,h^{E_1}\oplus h^{E_2})|_x&=\det\left(\mathrm{Id}_r+\frac{\sqrt{-1}}{2\pi}\begin{pmatrix} R^{E_1}_x&0 \\ 0& R^{E_2}_x \end{pmatrix}\right)\\ &=\det\left(\mathrm{Id}_r+\frac{\sqrt{-1}}{2\pi}\begin{pmatrix} R^{E}|_{E_{1,x}}&0 \\ 0& R^{E}|_{E_{2,x}} \end{pmatrix}\right)\\ &=\det\left(\mathrm{Id}_r+\frac{\sqrt{-1}}{2\pi}R^E_x\right)\\ &=c(E,h^E)|_x, \end{align*} $$

which follows that

$$ \begin{align*} P_\lambda(c(E,h^E))|_x=P_\lambda(c(E_1\oplus E_2,h^{E_1}\oplus h^{E_2}))|_x. \end{align*} $$

By Proposition 5.3, one has

(5.7) $$ \begin{align} P_\lambda(c(E,h^E))|_x&=\sum_{\mu,\nu}c^\lambda_{\mu\nu}P_\mu(c(E_1,h^{E_1}))|_x\wedge P_\nu(c(E_2,h^{E_2}))|_x. \end{align} $$

Since $(E_1,h^{E_1})$ is Nakano positive and $(E_2,h^{E_2})$ is dual Nakano positive at x, then $P_\mu (c(E_1,h^{E_1}))|_x$ and $P_\nu (c(E_2,h^{E_2}))|_x$ are positive forms. Since the Littlewood-Richardson coefficients $c^\lambda _{\mu \nu }$ are non-negative integers (see [Reference Fulton10, Corollary 1 in Chapter 5]), then each summand

$$ \begin{align*} c^\lambda_{\mu\nu}P_\mu(c(E_1,h^{E_1}))|_x\wedge P_\nu(c(E_2,h^{E_2}))|_x \end{align*} $$

in RHS of (5.7) is non-negative. However, for any $\lambda ,\mu ,\nu $ satisfying $\lambda _i=\mu _i+\nu _i$ for all i, then $c^{\lambda }_{\mu ,\nu }=1$ (see [Reference Fulton10, Page 66]), and

$$ \begin{align*}P_\mu(c(E_1,h^{E_1}))|_x\wedge P_\nu(c(E_2,h^{E_2}))|_x\end{align*} $$

is a positive $(|\lambda |,|\lambda |)$ -form by Proposition 3.4. By (5.7), we show that the Schur form $P_\lambda (c(E,h^E))|_x$ is a positive $(|\lambda |,|\lambda |)$ -form.