The set of strongly positive forms is a closed cone

Question

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.

Answer 1

The set $P$ of strongly positive forms is the convex hull of the cone of simple strongly positive forms which will be denoted by $$\mathcal{C}=\{\Omega\in\bigwedge^{p,p}V|\Omega=\sqrt{-1}^{p^{2}}w\wedge \overline{w}=\sqrt{-1}^{p^{2}}w_{1}\wedge\cdots\wedge w_{p}\wedge \overline{w_{1}} \wedge\cdots\wedge \overline{w_{p}}\}.$$ Suppose $\mathcal{C}$ is closed. Let $\mathcal{S}$ be the unit sphere of $\bigwedge^{p,p}V$. Then $K=\mathcal{C}\cap \mathcal{S}$ is compact. And $conv(K)$ which denotes the convex hull of $K$ is a compact set without origin. Suppose to the contrary, it contains the origin, then we have $\sum_{i}\lambda_{i}v_{i}=0$ with $\lambda_{i}>0$ and $v_{i}\in K$. Since $v_{i}$'s can be regraded as positive definite matrixes, this can not happen. Then we can use the compactness and origin freeness to show that it will generate a closed cone $G$. And it is obvious that $G=P$. Hence it suffices to show that $\mathcal{C}$ is closed.

Suppose we have a Cauchy sequence of simple forms $(\Omega_{i}=\sqrt{-1}^{p^{2}}w_{i}\wedge \overline{w_{i}})$, we need to show that its limit $\Omega$ remains in $\mathcal{C}$. If we fix a set of basis in $\bigwedge^{p,0}V$, then $w_{i}$ is realized as a column vector $x_{i}$ and $\overline{w_{i}}$ is realized as a row vector $x_{i}^{*}$. From this point of view, each $w_{i}\wedge \overline{w_{i}}$ is in one-to-one correspondence to a Hermitian matrix $A_{i}=x_{i}x_{i}^{*}$. Let's choose the operator norm for the space of Hermitian matrixes. Since the choice of norms does't influence the topology of a vector space, $A_{i}$ converges to some $A$ as $\Omega_{i}$ does. Moreover, with this norm at hand, it is easy to show that $\sup_{i}\left\| x_{i} \right\|^{2}\leq\sup_{i}\left\| A_{i} \right\|< \infty$. By compactness, $x_{i}$ converges to some $y$, and it is obvious that the limit matrix $A=yy^{*}$. Now it suffices to show that $y$ represents a decomposable element in $\bigwedge^{p,0}V$. But by Plücker embedding, the set of decomposable forms in $\bigwedge^{p,0}V$ is closed. So $y$ as the limit of a subsequence of decomposable elements $x_{i}$ is also decomposable.