Fredholm index

Let $P$ be a Fredholm operator. The index of $P$ is defined as

	$\displaystyle\mathop{\mathrm{index}}\nolimits(P)$	$\displaystyle=$	$\displaystyle\dim\ker(P)-\dim\mathop{\mathrm{coker}}\nolimits(P)$
		$\displaystyle=$	$\displaystyle\dim\ker(P)-\dim\ker(P^{*}).$

Note: this is well defined as $\ker(P)$ and $\ker(P^{*})$ are finite-dimensional vector spaces, for $P$ Fredholm.

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$\mathop{\mathrm{index}}\nolimits(P^{*})=-\mathop{\mathrm{index}}\nolimits(P)$ .
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$\mathop{\mathrm{index}}\nolimits(P+K)=\mathop{\mathrm{index}}\nolimits(P)$ for any compact operator $K$ .
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If $P_{1}\colon\mathord{\mathcal{H}}_{1}\to\mathord{\mathcal{H}}_{2}$ and $P_{2}\colon\mathord{\mathcal{H}}_{2}\to\mathord{\mathcal{H}}_{3}$ are Fredholm operators, then $\mathop{\mathrm{index}}\nolimits(P_{2}P_{1})=\mathop{\mathrm{index}}\nolimits(% P_{1})+\mathop{\mathrm{index}}\nolimits(P_{2})$ .
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If $t\to P_{t}$ , $t\in[0,1]$ is a norm continuous path of Fredholm operators, then $\mathop{\mathrm{index}}\nolimits(P_{t})=\mathop{\mathrm{index}}\nolimits(P_{0})$ .

Fredholm operators of the form $\mathit{invertible}+\mathit{compact}$ have index zero.