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What can be said about the tensor product $T\otimes S$ of two Fredholm operators $T:X_1\to Y_1$ and $S:X_2 \to Y_2$ where $X_1,X_2,Y_1, Y_2$ are Banach spaces and tensor product of operators are considered on the projective tensor product of underling Banach spaces? Is the index of tensor product operator $T\otimes S$ bounded by the index of $T$ and index of $S$? What about if we replace the projective tensor product with other types of Banach space tensor product norm?

As a particular case what is a precise formula for the tensor product of two Laplacian operators on a Riemannian manifold? Namely assume that we have two Riemannian metrics $g_1,g_2$ on $M$ with corresponding Laplacian $\Delta_{g_1}, \Delta_{g_2}$. Is $\Delta_{g_1}\otimes \Delta_{g_2}$ a Laplacian of a metric on $M$?

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    $\begingroup$ The tensor product $L^2(M,g_1) \otimes L^2(M,g_2)$ is canonically isomorphic to $L^2(M \times M,g_1 \oplus g_2)$, where, by abuse of notation, $g_1 \oplus g_2$ denotes the metric on $M \times M$ induced from $g_1$ and $g_2$ via the canonical isomorphism $T(M \times M) \cong \operatorname{Proj_1}^\ast TM \oplus \operatorname{Proj_2}^\ast TM$. Then $\Delta_{g_1 \oplus g_2}$ can identified with $\Delta_{g_1} \otimes I + I \otimes \Delta_{g_2}$, not $\Delta_{g_1} \otimes \Delta_{g_2}$. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 28 at 20:20
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    $\begingroup$ Why would $T \otimes S$ even be Fredholm? If $T$ has a nontrivial kernel and $\text{dim}(X_2) = \infty$, then $T \otimes S$ has infinite-dimensional kernel and so is not Fredholm, no? $\endgroup$
    – David Gao
    Commented Sep 28 at 20:22
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    $\begingroup$ @AliTaghavi That's impossible since $\Delta_{g_1} \otimes \Delta_{g_2}$ is a fourth-order [!!!] partial differential operator. For example, if $M = \mathbb{R}$ and $g_1 = g_2$ is the usual flat metric, then $\Delta_{g_1} = \Delta_{g_2} = -\tfrac{\mathrm{d}^2}{\mathrm{d}t^2}$ and $\Delta_{g_1} \otimes \Delta_{g_2}$ can be identified with $(-\partial_1^2)(-\partial_2^2) = \partial_1^2\partial_2^2$ on $M \times M = \mathbb{R}^2$ with the usual flat metric. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 28 at 21:03
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    $\begingroup$ The tensor product $L^2(M,g_1) \otimes L^2(M,g_2)$ is canonically isomorphic to $L^2(M \times M,g_1 \oplus g_2)$, so I'm afraid I don't understand what you're hoping for. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 28 at 21:16
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    $\begingroup$ Because your question involves a tensor product over the scalars, not a balanced tensor product. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 29 at 11:51

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