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?

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1
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. – Sanchez Jan 18 at 8:37

2
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 at 11:33
@DavidLoeffler, sorry, I will pay attention to it. – hxhxhx88 Jan 18 at 16:05
@Sanchez, maybe I will have once more try, thanks. – hxhxhx88 Jan 18 at 16:06
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