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This question comes from the Complex Analytic and Differential Geometry by Demailly. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if $$u= \sum_{s=1}^N \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.

Here is my question: Demailly claimed that the set of strongly positive forms is a closed set and he didn't give a proof. I don't think it is an obvious result: Consider an absolutely convergent infnite sum of $(q,q)$-forms: $$\sum_{s} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$$$\sum_{s=1}^{\infty} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ It is an element in the closure of strongly positive cone, but I can't prove it is strongly positive unless $p=1,n-1$.

There is a similar (but not the same) question on math Stack Exchange: question. So I copy part of that question here for convenience.

This question comes from the Complex Analytic and Differential Geometry by Demailly. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if $$u= \sum_{s=1}^N \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.

Here is my question: Demailly claimed that the set of strongly positive forms is a closed set and he didn't give a proof. I don't think it is an obvious result: Consider an absolutely convergent infnite sum of $(q,q)$-forms: $$\sum_{s} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ It is an element in the closure of strongly positive cone, but I can't prove it is strongly positive unless $p=1,n-1$.

There is a similar (but not the same) question on math Stack Exchange: question. So I copy part of that question here for convenience.

This question comes from the Complex Analytic and Differential Geometry by Demailly. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if $$u= \sum_{s=1}^N \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.

Here is my question: Demailly claimed that the set of strongly positive forms is a closed set and he didn't give a proof. I don't think it is an obvious result: Consider an absolutely convergent infnite sum of $(q,q)$-forms: $$\sum_{s=1}^{\infty} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ It is an element in the closure of strongly positive cone, but I can't prove it is strongly positive unless $p=1,n-1$.

There is a similar (but not the same) question on math Stack Exchange: question. So I copy part of that question here for convenience.

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The set of strongly positive forms is a closed cone

This question comes from the Complex Analytic and Differential Geometry by Demailly. Let $V$ be a $n$ dimensional complex space. Consider the exterior algebra $\Lambda V^* = \oplus \Lambda^{(p,q)}V^*$. $u \in \Lambda^{(q,q)}V^*$ is called strongly positive if $$u= \sum_{s=1}^N \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ where $\gamma_s \ge 0$. $v \in \Lambda^{(p,p)}V^*$ where $p+q = n$ is called positive if $u\wedge v$ is positive $(n,n)$-form, i.e. $u\wedge v = \lambda i\mathrm{d}z_1 \wedge \mathrm{d}\bar{z}_1 \wedge \cdots \wedge i\mathrm{d}z_p \wedge \mathrm{d}\bar{z}_p$ where $\lambda$ is a real positive number, for all $u$ strongly positive.

Here is my question: Demailly claimed that the set of strongly positive forms is a closed set and he didn't give a proof. I don't think it is an obvious result: Consider an absolutely convergent infnite sum of $(q,q)$-forms: $$\sum_{s} \gamma_s i\alpha_{s,1} \wedge \bar{\alpha}_{s,1} \wedge \cdots \wedge i\alpha_{s,q}\wedge \bar{\alpha}_{s,q}$$ It is an element in the closure of strongly positive cone, but I can't prove it is strongly positive unless $p=1,n-1$.

There is a similar (but not the same) question on math Stack Exchange: question. So I copy part of that question here for convenience.