I have some trouble in the definition of the hermitian metric on a hermitian manifold:

Let $M$ be a riemannian manifold of real dimension $n>1$ with real coordinates $x_1,\dots,x_n$. $M$ has a smoothly varying bilinear form on each tangent space, $ds^2=\sum_{i,j}g_{ij}dx_i\otimes dx_j$. Now suppose that $ds^2$ is the standard bilinear form of $\mathbb{R}^n$, then we can write $ds^2=\sum_{i=1}^ndx_i\otimes dx_i$ and it's such that, given the tangent vector $v=\sum_{j=1}^nv_i\frac{\partial}{\partial x_j}$, $ds^2(v,v)=\sum_{i=1}^ndx_i(\sum_{j=1}^nv_i\frac{\partial}{\partial x_j})\otimes dx_i(\sum_{j=1}^nv_i\frac{\partial}{\partial x_j})=\sum_{i=1}^nv_i^2$. I can't do this same thing on a hermitian manifold:

Let $X$ be a hermitian manifold of complex dimension $k>1$ with complex coordinates $z_1,\dots,z_k$. $X$ has a smoothly varying hermitian metric $h=\sum_{i,j}h_{i,j}dz_i\otimes d\overline{z_j}$ on each tangent space. As before, suppose that $h$ is the standard hermitian metric of $\mathbb{C}^k$, then $h=\sum_{i=1}^kdz_i\otimes d\overline{z_i}$. Given a tangent vector $w=\sum_{j=1}^kw_j\frac{\partial}{\partial z_j}$ it should be of course $h(w,w)=\sum_{j=1}^kw_j\overline{w_j}=\sum_{j=1}^k|w_j|^2$. But what I get is $h(w,w)=\sum_{i=1}^kdz_i(\sum_{j=1}^kw_j\frac{\partial}{\partial z_j})\otimes d\overline{z_i}(\sum_{j=1}^kw_j\frac{\partial}{\partial z_j})=0$ since $d\overline{z_i}(\frac{\partial}{\partial z_i})=\frac 1 2(dx_i-idy_i)(\frac{\partial}{\partial x_i}-i\frac{\partial}{\partial y_i})=0$.

There must be an error, can someone explain it to me?

Also, I don't understand what do the conjugate tangent vectors $\frac{\partial}{\partial \overline{z_j}}$ stand for. What should be the result of $h( \frac{\partial}{\partial \overline{z_j}},\frac{\partial}{\partial \overline{z_j}})$ and $h(\frac{\partial}{\partial z_j},\frac{\partial}{\partial \overline{z_j}})$ if $h$ is the standard hermitian metric?


1 Answer 1


$\newcommand{\dd}{\partial}$If I understand your question, the essential issue arises already on a complex curve. For simplicity, then, write a local complex coordinate as $z = x + iy$ with $x$ and $y$ real, so that $$ \begin{aligned} dz &= dx + i\, dy,\vphantom{\frac{\dd}{\dd}} \\ d\bar{z} &= dx - i\, dy,\vphantom{\frac{\dd}{\dd}} \end{aligned} \qquad \begin{aligned} \dd_{z} &:= \frac{\dd}{\dd z} = \frac{1}{2}\left[\frac{\dd}{\dd x} - i\, \frac{\dd}{\dd y}\right], \\ \dd_{\bar{z}} &:= \frac{\dd}{\dd \bar{z}} = \frac{1}{2}\left[\frac{\dd}{\dd x} + i\, \frac{\dd}{\dd y}\right]. \end{aligned} $$ It's easy to check that $$ dz(\dd_{z}) = d\bar{z}(\dd_{\bar{z}}) = 1,\qquad dz(\dd_{\bar{z}}) = d\bar{z}(\dd_{z}) = 0, $$ and that $$ dz \otimes d\bar{z} = (dx \otimes dx + dy \otimes dy) - i(dx \otimes dy - dy \otimes dx). $$ Consequently, $$ dz \otimes d\bar{z}(\dd_{z}, \dd_{\bar{z}}) = 1,\qquad dz \otimes d\bar{z}(\dd_{z}, \dd_{z}) = dz \otimes d\bar{z}(\dd_{\bar{z}}, \dd_{z}) = dz \otimes d\bar{z}(\dd_{\bar{z}}, \dd_{\bar{z}}) = 0. $$ Particularly, $(dz \otimes d\bar{z})(w, \bar{w}) = |w|^{2}$ for every $(1, 0)$-vector $w$, while as you note, $(dz \otimes d\bar{z})(w, w) = 0$.

  • $\begingroup$ but shouldn't it be $h(w,w)=|w|^2$ if $h$ is the standard hermitian metric of $\mathbb{C}$? $\endgroup$
    – user55835
    Commented Sep 20, 2016 at 12:12
  • 1
    $\begingroup$ Actually not. In fancy language, a Riemannian metric $g$ on a complex manifold $M$ has a unique complex-bilinear extension to the complexified tangent bundle of $M$. The spaces of $(1, 0)$- and $(0, 1)$-vectors are isotropic for the extension (as your calculation shows). To get an Hermitian metric on the "holomorphic tangent bundle" of $(1, 0)$-vectors, one puts $$h(X, Y) = g(X, \bar{Y}).$$ $\endgroup$ Commented Sep 20, 2016 at 12:23
  • $\begingroup$ @AndrewD.Hwang If I understand this correctly, $h$ is not the $dz\otimes d\bar{z}$ thing? $\endgroup$ Commented Sep 20, 2016 at 14:56
  • $\begingroup$ @PeterFranek: OP uses $h$ as both a general Hermitian metric and the flat metric. The main point of the question, however, appears to be the reason why a metric vanishes on a pair of $(1, 0)$-vectors, for which the metric components are effectively immaterial. $\endgroup$ Commented Sep 20, 2016 at 15:51
  • 1
    $\begingroup$ Ah, ok, I somehow still thought that $h(w,w)=|w|^2$ but now I see that you don't claim this. I deleted my previous comment.Thanks! $\endgroup$ Commented Sep 20, 2016 at 22:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .