# Automorphism on a kahlerian variety which acts as -id on $H^2(X,\mathbb{Z})$

I've read this proposition:

"there is no automorphism (biholomorphic map) of a Kahlerian complex manifold $X$ which acts on $H^2(X,\mathbb{Z})$ as $-$identity"

so I tried to prove this statement and i would like to ask you if I'm correct. Here is as I proceed:

Suppose there is such an automorphism, call it $f$. Then $f^*$ brings a Kahler class $\omega$ to $-\omega$. Writing $\omega$ as $g(I-,-)$, where $g$ is a kahlerian metric, this means that $f^*$ brings $g(I-,-)$ to $-g(I-,-)$. That means that on the second copy of $X$ there should be a riemannian metric $-g$, but it is not an inner product on the tangent bundle (for the positive condition).

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How are you explicitly using the fact that $f$ is holomorphic? – Ted Shifrin Jul 20 '13 at 23:46
i realized i didn't specify that $I$ is the complex structure.. anyway seeing $I$ as an integrable almost complex structure, the fact that $f$ is holomorphic means that $f_*I=If_*$ on the tangent spaces $T_pX$. So $f_*$ must bring a $(1,0)$ tangent vector to a $(1,0)$ tangent vector and so the minus sign must go on the riemannian metric. – jim kriegsmann Jul 20 '13 at 23:54
OK. I prefer to think about the fact that $\omega^k/k!$ is the induced volume form on $k$-dimensional complex submanifolds. – Ted Shifrin Jul 21 '13 at 0:11
well.. but wouldn't it work only if the complex dimension $k$ is odd? because if it's even then $f^*$ sends the volume form to itself and there is no problem in that – jim kriegsmann Jul 21 '13 at 0:15
Yup, sure. But apply it with $k=1$ :) – Ted Shifrin Jul 21 '13 at 0:17