# Index of tensor product of two Fredholm operators

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 $$j(T)=\operatorname{dim}(\ker(T))-\operatorname{dim}(\ker(T^{*}))$$.

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.