Timeline for The tensor product of two Fredholm operators
Current License: CC BY-SA 4.0
24 events
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| Oct 1 at 6:32 | comment | added | David Gao | @AliTaghavi Yes. Unless $T$ and $S$ are both bijective, or if some of the spaces involved are finite-dimensional, $T \otimes S$ can never be Fredholm. | |
| Sep 30 at 4:32 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Fixed typos
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| Sep 30 at 0:52 | review | Close votes | |||
| Oct 9 at 3:05 | |||||
| Sep 29 at 11:51 | comment | added | Branimir Ćaćić | Because your question involves a tensor product over the scalars, not a balanced tensor product. | |
| Sep 29 at 7:18 | comment | added | Ali Taghavi | @BranimirĆaćić Yes I see in fact the problem is that the Laplacian is not $C^{\infty}(M) $ linear. However the tensor product in your first comment is taken over scalars. | |
| Sep 28 at 21:43 | comment | added | Branimir Ćaćić | It follows from Serre–Swan that if $E$ and $F$ are Hermitian vector bundles over $M$ compact, then $L^2(M,E \otimes F) \cong \Gamma_{\mathrm{cts}}(M,E) \otimes_{C(M)} L^2(M,F)$, which is a balanced tensor product over the $C^\ast$-algebra $C(M)$. You may be thinking of the process of twisting a Dirac-type operator by a Hermitian vector bundle with connection, which does not involve producing a new Fredholm operator via a straightforward tensor product of Fredholm operators. | |
| Sep 28 at 21:31 | comment | added | Ali Taghavi | So $x^6\otimes 1$ is maped to 0 but $x^3\otimes x^3$ is maped to $6x\otimes 6x$ identified to $36x^2$. I think I am doing a stupid mistake but what is my mistake?(I repeat that I do not product M with M. I just consider the tensor trivial bundle and the tensor operrator. | |
| Sep 28 at 21:28 | comment | added | Ali Taghavi | @BranimirĆaćić Sorry if the point is trivial: where is the contraditory point: The trivial line bundle on R is denoted by $\epsilon_1$ then $\epsilon_1\otimes \epsilon_1 \sim \epsilon_1$ then the section space of tensor product is isomorphisc to the tensor product of the corresponding sectioin(Serre Swan theorem). On the other hand the isomorphisms between $\mathbb{R}\otimes \mathbb{R}$ with $\mathbb{R}$ is in the form $a\otimes b \mapsto ab$. Now we wish to look at $\Delta \otimes \Delta$ acted on , say $x^6$... ok.. | |
| Sep 28 at 21:16 | comment | added | Branimir Ćaćić | 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. | |
| Sep 28 at 21:13 | comment | added | Ali Taghavi | @BranimirĆaćić Yes but in the post I did not product manifolds. The tensor product of $\epsilon_1$ with itself is the trivial line bundle on $M$(The base space is always $M$. so $f(x)\otimes g(x)$ is identified with $f(x)g(x)$ So I guess I am leading to triviality!! because $f\otimes 1$ maps to 0 !! am Imissing some thing? how can I improve the question in Riemannian metric case? | |
| Sep 28 at 21:03 | comment | added | Branimir Ćaćić | @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. | |
| Sep 28 at 20:51 | comment | added | Ali Taghavi | I mean module constant functions | |
| Sep 28 at 20:39 | comment | added | Ali Taghavi | @DavidGao Very good point. A similar point for codimension? | |
| Sep 28 at 20:38 | comment | added | Ali Taghavi | @BranimirĆaćić But very good point you indicated to | |
| Sep 28 at 20:37 | comment | added | Ali Taghavi | @BranimirĆaćić ..possible other metrics | |
| Sep 28 at 20:36 | comment | added | Ali Taghavi | So in the compact case there is no any harmonic map so there is a chance of ellupticity (since there is no immediate obstruction for Fredholm ness of tensor product of laplacians). Yes the Laplacian of product metric is not the tensor product of corresponding Laplaciqn but what about possible... | |
| Sep 28 at 20:31 | comment | added | Ali Taghavi | @DavidGao Thanks for your comment | |
| Sep 28 at 20:31 | comment | added | Ali Taghavi | @BranimirĆaćić Thanks for your comment | |
| Sep 28 at 20:22 | comment | added | David Gao | 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? | |
| Sep 28 at 20:20 | comment | added | Branimir Ćaćić | 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 $\Delta_{g_1} \otimes \Delta_{g_2}$. | |
| Sep 28 at 20:06 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 28 at 19:33 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 28 at 19:11 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 28 at 19:05 | history | asked | Ali Taghavi | CC BY-SA 4.0 |