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A real $(p,p)$-form $\Omega$ on complex manifold $X$ is called strictly positive $(p,p)$-form if $$(-\sqrt{-1})^{p}\Omega(v_1,\bar{v}_1,...,v_p,\bar{v}_p)>0$$ for any nonzero $v_1,v_2,...,v_p\in T^{1,0}(X)$.If $\omega$ its Hermitian form.

Does the $\omega^p$ is strictly positive $(p,p)$-form?

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2 Answers 2

Actually, you don't have definition quite right. You have to assume that $v_1,v_2,\ldots,v_p$ are linearly independent over $\mathbb{C}$, not just nonzero, otherwise the quantity you have displayed will certainly be zero.

With this corrected definition, though, one clearly does have $\omega^p$ is positive for any real, positive $(1,1)$-form $\omega$ on a complex manifold. This follows immediately, for example, from the fact that the symmetry group of a positive Hermitian inner product on a complex vector space $V$ acts transitively on the set of complex $p$-planes in $V$.

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I will try to clarify a little bit your question and at the same time to give you an answer. All that I say, you can find on Demailly's book "Complex Analytic and Differential Geometry".

GENERAL THEORY

Let $V$ be a complex vector space of dimension $n$ and $(z_1,\dots,z_n)$ coordinates on $V$. We denote by $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ the corresponding basis of $V$, by $(dz_1,\dots,dz_n)$ its dual basis in $V^*$ and consider the exterior algebra $$ \Lambda V^*_\mathbb C=\bigoplus\Lambda^{p,q}V^*,\quad \Lambda^{p,q}V^*=\Lambda^p V^*\otimes\Lambda^q\overline{V^*}. $$ Let us first observe that V has a canonical orientation, given by the $(n, n)$-form $$ \tau(z)=idz\wedge d\bar z_1\wedge\cdots\wedge idz_n\wedge d\bar z_n=2^n dx_1\wedge dy_1\wedge\cdots\wedge dx_n\wedge dy_n, $$ where $z_j=x_j+iy_j$. In fact, if $(w_1,\dots,w_n)$ are other coordinates, we find $$ dw_1\wedge\cdots\wedge d w_n=\det(\partial w_j/\partial z_k)dz_1\wedge\cdots\wedge dz_n, $$ $$ \tau(w)=|\det(\partial w_j/\partial z_k)|^2\tau(z). $$ Definition. A $(p,p)$-form $u\in\Lambda^{p,p}V^*$ is said to be positive if for all $\alpha_j\in V^*$, $1\le j\le q=n-p$, then $$ u\wedge i\alpha_1\wedge\bar\alpha_1\wedge\cdots\wedge i\alpha_q\wedge\bar\alpha_q $$ is a positive $(n,n)$-form. A $(q,q)$-form $v\in\Lambda^{q,q}V^*$ is said to be strongly positive if $v$ is a convex combination $$ v=\sum\gamma_si\alpha_{s,1}\wedge\bar\alpha_{s,1}\wedge\cdots\wedge i\alpha_{s,q}\wedge\bar\alpha_{s,q} $$ where $\alpha_{j,s}\in V^*$ and $\gamma_s\ge 0$.

It is straightforward to see that strongly positive implies positive and, moreover, the concepts of positive and strongly positive coincide in bidegree $(0,0)$, $(1,1)$, $(n-1,n-1)$ and $(n,n)$.

Now, you have that all positive form $u$ are real, that is satisfy $u=\bar u$. In particular, in terms of coordinates, if $$ u=i^{p^2}\sum_{|I|=|J|=p}u_{I,J}dz_I\wedge d\bar z_J, $$ then the coefficients satisfy the hermitian symmetry relation $\overline u_{I,J}=u_{J,I}$.

A form $u=i\sum_{j,k}u_{jk}dz_j\wedge d\bar z_k$ of bidegrees $(1,1)$ is positive if and only if $\xi\mapsto\sum u_{jk}\xi_j\bar\xi_k$ is a semi-positive hermitian form on $\mathbb C^n$.

Proposition. If $u_1,\dots,u_s$ are positive forms, all of them strongly positive (resp. all except perhaps one), then $u_1\wedge\cdots\wedge u_s$ is strongly positive (resp. positive).

Proof. The proof is immediate from the very definition.

What you are asking follows straightforwardly from what I have written (in particular the last proposition), taking all $u_j=\omega$.

WITHOUT GENERAL THEORY

If $\omega$ comes from a (positive definite) hermitian metric, then you can always choose coordinates such that your form $\omega$ is given by $\omega=i\sum_{j=1}^n dz_j\wedge d\bar z_j$. Then $$ \omega^p= i^{p^2}p!\sum_{|I|=p}dz_I\wedge d\bar z_I. $$ I leave you the pleasure to compute $\omega^p(v_1,\dots,v_p,\bar v_1,\dots,\bar v_p)$ and to discover how this quantity is related with the determinants of the order $p$ minors of the matrix whose columns are given by $v_1,\dots,v_p$.

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Thanks dear diverietti,Robert Bryant –  Gran Murra Dec 26 '11 at 0:54

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