Let $S$ and $T$ Fredholm operators on a separable complex Hilbert space $\mathcal H$ such that the tensor product $S\otimes T$ is also a Fredholm operator on $\mathcal H\otimes\mathcal H$. So what to say about the index: $$ j(S\otimes T)=? $$ Remember that $j(T)=\operatorname{dim}(\ker(T))-\operatorname{dim}(\ker(T^{*}))$.
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2 In the finite-dimensional setting, every operator is Fredholm with index zero. So let's assume $\mathcal H$ is infinite-dimensional.
Firstly, note that if $x\in\ker(T)$ is non-trivial, then $x\otimes y\in\ker(T\otimes S)$ for any $y\in\mathcal H$. Thus $\ker(T\otimes S)$ is infinite-dimensional as long as either $\ker(T)$ or $\ker(S)$ is non-trivial, and therefore $T\otimes S$ is not Fredholm. Similarly, we see that if $\ker(T^*)$ or $\ker(S^*)$ is non-trivial, then $T\otimes S$ is not Fredholm. Thus the only time that $T\otimes S$ is Fredholm is when $T$ and $S$ are both invertible, in which case $T\otimes S$ is invertible, and the index of all three operators is zero.
- $\begingroup$ Thank you very much. $\endgroup$Marcos Ferreira– Marcos Ferreira2019-07-15 23:42:43 +00:00Commented Jul 15, 2019 at 23:42
- $\begingroup$ No problem. Glad to help! $\endgroup$Aweygan– Aweygan2019-07-15 23:58:41 +00:00Commented Jul 15, 2019 at 23:58