# Why is the Petersson inner product positive definite?

The Petersson inner product is defined on the space $\mathcal{S}_k(\Gamma)$ of weight $k$ cusp forms of level $\Gamma$, and takes values in $\mathbb{C}$.

First of all, I wonder: what does it mean for a complex-valued inner product to be positive definite?

Then, can anyone show me why the Petersson inner product is positive definite?

-
Positive definite means that $\langle x,x \rangle \ge 0$ for all $x$ and equality holds iff $x = 0$. That Petersson inner product is positive definite should be self evident. –  user27126 Jan 18 '13 at 8:37
I cleaned up the spelling and grammar in your post. If you're expecting people to take the trouble to answer your question, you should take a little more trouble over asking it. –  David Loeffler Jan 18 '13 at 11:33
@DavidLoeffler, sorry, I will pay attention to it. –  hxhxhx88 Jan 18 '13 at 16:05
@Sanchez, maybe I will have once more try, thanks. –  hxhxhx88 Jan 18 '13 at 16:06

A Hermitian form $\langle \cdot,\cdot\rangle$ is positive definite if $\langle f,f\rangle >0$ whenever $f\ne 0$.
Because plugging $g=f$ in the definition, we get an integral of a nonnegative function (which is zero a.e. if and only if $f=0$ a.e.). One should worry a little about the integral being convergent and appropriately behaved under the action of the modular group, but this is really the issue of the product being well-defined rather than positive-definite. You can find the details on page 74-75 of Introduction to the Arithmetic Theory of Automorphic Functions by Shimura.