Timeline for Injection vs. bijection between $V^\star \otimes V^\star$ and $(V \otimes V)^\star$ [closed]
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23 events
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| Apr 20, 2022 at 21:22 | history | edited | MathTolliob | CC BY-SA 4.0 |
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| Apr 14, 2022 at 20:11 | history | closed |
LSpice Najib Idrissi abx Andreas Blass John Wiltshire-Gordon |
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| S Apr 14, 2022 at 19:57 | history | suggested | J. W. Tanner | CC BY-SA 4.0 |
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| Apr 14, 2022 at 19:19 | review | Suggested edits | |||
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| Apr 14, 2022 at 15:29 | comment | added | YCor | One-to-one correspondence (as opposed to "one-to-one map") sounds like a bijection, so I'd rather rephrase the title as "Injection vs bijection..." | |
| Apr 14, 2022 at 15:28 | history | edited | YCor | CC BY-SA 4.0 |
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| Apr 14, 2022 at 14:35 | review | Close votes | |||
| Apr 14, 2022 at 20:11 | |||||
| Apr 14, 2022 at 14:32 | comment | added | Najib Idrissi | This isn't really a research-level question. Dozens of variations have been asked and answered on MSE, see e.g. math.stackexchange.com/q/679584/10014 math.stackexchange.com/q/573378/10014 | |
| Apr 14, 2022 at 13:56 | comment | added | R. van Dobben de Bruyn | Whereas when $V$ is of infinite dimension, the two sides are always abstractly isomorphic: if $\dim V$ is some infinite cardinal $\kappa$, then $V \otimes V$ has dimension $\kappa^2 = \kappa$, hence $V \cong V \otimes V$. Similarly, $V^* \otimes V^* \cong V^*$, so both sides are (very non-canonically) isomorphic to $V^*$. | |
| Apr 14, 2022 at 13:52 | answer | added | YCor | timeline score: 9 | |
| Apr 14, 2022 at 13:43 | comment | added | MathTolliob | @R.vanDobbendeBruyn When $V$ is of infinite dimension, the natural map is never an isomorphism, otherwise, the element constructed by Wojowu would have been in the image of $V^\star \otimes V^\star$ and would not have been a counter-example | |
| Apr 14, 2022 at 13:42 | comment | added | YCor | @NikWeaver this is not a basis. | |
| Apr 14, 2022 at 13:39 | history | edited | MathTolliob | CC BY-SA 4.0 |
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| Apr 14, 2022 at 13:37 | comment | added | MathTolliob | @NikWeaver What you said is not clear for me... Could you, please, explicit it? | |
| Apr 14, 2022 at 13:35 | comment | added | MathTolliob | @R.vanDobbendeBruyn You're totaly right! That is my question. An answer should be constructed from Wojowu counter example | |
| Apr 14, 2022 at 13:32 | history | edited | YCor |
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| Apr 14, 2022 at 12:57 | comment | added | R. van Dobben de Bruyn | I think the final question is not quite what you want to ask. It's likely that in many (all?) cases, the dimensions of $V^* \otimes V^*$ and $(V \otimes V)^*$ are the same, so they are abstractly isomorphic. That's not the interesting question; the interesting question is whether the natural map $V^* \otimes V^* \to (V \otimes V)^*$ is an isomorphism. | |
| Apr 14, 2022 at 12:52 | comment | added | Wojowu | @NikWeaver You can identify the algebraic dual with the set of functions from $B$ into $k$, but how do you form a basis out of it? | |
| Apr 14, 2022 at 11:12 | comment | added | MathTolliob | Could we nevertheless construct a basis of $(V \otimes V)^\star$? | |
| Apr 14, 2022 at 11:09 | comment | added | MathTolliob | OK, nice (counter-)example! The problem comes from the fact that this element could not be spanned by a finite sum of $(b_1 \otimes b_2)^\star$... Thank you Wojowu | |
| Apr 14, 2022 at 10:52 | comment | added | Wojowu | $B^*$ does not span $V^*$ if $V$ is infinite-dimensional, and similarly for $C^*$. As for an "explicit example", literally any infinite-dimensional $V$: formally, we can define an element of $(V\otimes V)^*$ which maps any element to the sum of coefficients in the expansion in the basis $b_1\otimes b_2$. | |
| Apr 14, 2022 at 10:48 | history | edited | MathTolliob | CC BY-SA 4.0 |
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| Apr 14, 2022 at 10:42 | history | asked | MathTolliob | CC BY-SA 4.0 |