3
$\begingroup$

I am reading Kobayashi book entitled Differential Geometry of Complex Vector bundles.

He introduced the concept of hermitian metric on a complex vector bundle $E \to M$, with $M$ not necessarily a complex manifold, being the tensor field such that at every point it defines a complex hermitian inner product.

So, if locally there exists a local frame $s = (s_1,\ldots,s_r)$ (Assuming rank $r$) and being $h$ such tensor field, I can easily compute the matrix associated to $h$ as:

$$h_{ij} := h(s_i,s_j).$$

In particular, the same procedure applies to holomorphic vector bundles, where the local frame has the property of being a holomorphic function between the basis manifold and the bundle, where the basis manifold and the total space are complex manifolds.

Considering $M$ being a complex $n-$dimensional manifold, the tangent bundle $TM$ to $M$ can be seen as a holomorphic vector bundle. In fact, if we consider $TM_{\mathbb{C}} := TM\otimes_{\mathbb{R}}\mathbb{C}$ then it splits as $$TM_{\mathbb{C}} = TM'\oplus TM'',$$ where the spaces on the decomposition are the eigenspaces associated to the extension to $TM_{\mathbb{C}}$ of the standard complex structure on $M$.

In local coordinates $(z^ 1,\ldots,z^ n)$ the space $TM'$ can be seen as the space generated by $\{\frac{\partial}{\partial z^ k}\}_{k =1}^ n.$ In particular, $TM$ possess the structure of a $n-$vector bundle once $TM \cong T'M.$

Here is where I get stuck. The author claims:

Let $TM$ as above with a hermitian metric $h$. Then, locally we can write the components of $h$ as: $$h_{i\overline{j}} := h(\frac{\partial}{\partial z^ i}, \frac{\partial}{\partial \overline{z}^ j}).$$

It is known that a complex manifold with a Riemannian metric $g$ orthogonal with the respect to the standard complex structure $J$ is called an hermitian manifold with hermitian metric $g$. The metric $g$ can be extended to $T_\mathbb{C}M$ and its only non-null components are precisely $h_{i\overline{j}}$ defined above.

So my question is: Why if we consider $TM$ as holomorphic vector bundle the expression for these two metrics coincide? Why are the mixed terms the only non-null terms even if we consider $TM$ as an holomorphic vector bundle? It should not appear any mixed index on the metric.. What am I missing?

$\endgroup$
2
$\begingroup$

Since the problem has a local nature, I'll consider only the Linear Algebra couterpart. Let $V$ be a real vector space whose dimension is $2n$ and let $$J\colon V \longrightarrow V$$ be a complex structure on $V$, $J^2 = -Id$. Then $(V,J)$ is a (n-dimensional) complex vector space for the scalar multiplication $$(a+bi)v = av+bJ(v).$$ Let $\left<,\right>$ be a (real) inner product on $V$ compatible with $J$, $$\left<J(v),J(w)\right> = \left<v,w\right>.$$ We define an hermitian product on $(V,J)$: $$\left(v,w\right) := \left<v,w\right> -i\left<J(v),w\right>.$$ On the other hand, we have a natural hermitian extension for $\left<,\right>$ to $V\otimes\mathbb{C}$: $$\left<v\otimes a,w \otimes b\right>_{\mathbb{C}} := a\bar b \left<v,w\right>.$$ Extending $J$ to $V\otimes\mathbb{C}$ we may take the eingenspaces for $J$ and make the decomposition $$V\otimes\mathbb{C} = V'\oplus V''.$$ Fix a (real) $\mathbb{C}-$basis $\{x_1, \dots x_n\}$ for $(V,J)$. Then the $y_j := J( x_j)$ complete a real basis for $V$. Define $$z_j = \frac{1}{2}(x_j -iy_j).$$ Since $J(z_j) =iz_j$, the set $\{z_1, \dots z_n\}$ is a $\mathbb{C}-$basis for $(V',i)$, the set $\{\bar z_1, \dots \bar z_n\}$ is a $\mathbb{C}-$basis for $(V'',-i)$ and $x_j \mapsto z_j$ , $x_j \mapsto \bar z_j$ are isomorphisms. Now we head to the product.

$$ \left<z_j,z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> +i \left<x_j,y_k\right> +\left<y_j,y_k\right> \right] = \frac{1}{2}\left(x_j, x_k\right), $$ $$ \left<z_j,\bar z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> -i \left<x_j,y_k\right> -\left<y_j,y_k\right> \right] = 0, $$ $$ \left<\bar z_j,\bar z_k\right>_{\mathbb{C}} = \overline{\left<z_j,z_k\right>}_{\mathbb{C}}. $$ We have then the relation between the hermitian products $\left(,\right)$ on $(V,J)$ and $\left<,\right>_{\mathbb{C}}$ on $(V\otimes \mathbb{C},i)$. The last thing to remark is that on Kobayashi's notation $$ h(z_j,z_k) = \left<z_j,\bar z_k\right>_{\mathbb{C}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.