Timeline for Examples of common false beliefs in mathematics
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 16 at 15:29 | comment | added | Todd Trimble | @მამუკაჯიბლაძე To keep things simple and manageable, I'm relying on the fact that every (symmetric) monoidal category is (symmetric) monoidally equivalent to a (symmetric) strict monoidal category, so that I may assume WLOG that $M$ is symmetric strict monoidal. Does this satisfactorily address your question? For some technical details on monoidal strictification, I might point to this section in the nLab, ncatlab.org/nlab/show/clique#monoidal_strictifications. Anyway, to be more accurate, the permutation groupoid $S$ is the free symmetric strict monoidal category on one generator. | |
| Nov 15 at 18:24 | comment | added | მამუკა ჯიბლაძე | @ToddTrimble Cool! But I am curious about details, e. g. it seems there are $n$th Catalan number many functors $M^n\to M$ in the image of the functor, with $S_n$ acting freely on homs between these? Is this written somewhere? | |
| Nov 15 at 15:22 | comment | added | Todd Trimble | @მამუკაჯიბლაძე "When they do coincide there also is the left action that I had in mind but maybe it can also be defined abstractly? Say, in any symmetric monoidal category?" <-- yes. For a symmetric monoidal category $M$, there is a sm category $M^\bullet$ whose objects are functors $M^n \to M$, and whose monoidal product is the obvious composition $M^m \times M^n \to M \times M \to M$. This has a distinguished object $1_M: M \to M$. The permutation groupoid $S$ is the free sm category on one object x. The induced sm functor $S \to M^\bullet$ taking $x$ to $1_M$ gives the desired left actions. | |
| Feb 12, 2021 at 8:52 | comment | added | მამუკა ჯიბლაძე | Whereas the left action operates on the orbit of $(\pi_1,...,\pi_n)\in\hom(X^n,X)^n\approx\hom(X^n,X^n)$ | |
| Feb 12, 2021 at 6:28 | comment | added | მამუკა ჯიბლაძე | @R.vanDobbendeBruyn Yes, in fact in any category (with products) it makes sense to view $X^n$ as "the object of maps from $\{1,...,n\}$ to $X$", on which automorphisms of $\{1,...,n\}$ act contravariantly. | |
| Feb 10, 2021 at 1:38 | comment | added | R. van Dobben de Bruyn | The same already happens with the permutation action of $S_n$ on $k^n$: the action $\sigma \cdot (x_1,\ldots,x_n) = (x_{\sigma(1)},\ldots,x_{\sigma(n)})$ is a right action. The better definition is $(\sigma x)_i = x_{\sigma(i)}$; the confusion comes from viewing coordinates as maps into versus maps out of your space. | |
| Feb 9, 2021 at 9:45 | comment | added | მამუკა ჯიბლაძე | Thank you! I overlooked your statement that the action is from the right, and was thinking about the left action that permutes the vectors according to $\sigma$. It is true that for different spaces $V_1$, ..., $V_n$ the simplest things available are canonical isomorphisms $V_1\otimes...\otimes V_n\to V_{\sigma(1)}\otimes...\otimes V_{\sigma(n)}$ and this gives that right action when the spaces coincide. When they do coincide there also is the left action that I had in mind but maybe it can also be defined abstractly? Say, in any symmetric monoidal category? | |
| Feb 9, 2021 at 8:15 | comment | added | Ivan Solonenko | @მამუკა ჯიბლაძე, one's mind gets tricked because in the original tensor, for each $v_i$, its subscript number coincides with its place within the tensor. And the action of $S_n$ is unfortunately formulated in terms of the subscripts. If you look at the definition of the action, it actually says: $\sigma$ takes a decomposable tensor and sends its $i$-th factor to the $\sigma^{-1}(i)$-th place. This inverse explains why the action is actually from the right. | |
| Feb 9, 2021 at 6:05 | comment | added | მამუკა ჯიბლაძე | Probably I will now demonstrate another false belief, but how is this possible? A $\tau$ is presented with an $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ to act on. It is not told what $\sigma$ was, how can it guess? In other words, this $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ is just some (rank 1) tensor, and $\tau$ is just supposed to permute its entries, without knowing anything about any previous actions on it, no? | |
| Feb 7, 2021 at 19:24 | comment | added | Simon Wadsley | This is something I have reteach myself every time I teach it to others. Happily I remember that I need to do so. | |
| Feb 7, 2021 at 17:45 | history | edited | Ivan Solonenko | CC BY-SA 4.0 |
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| Feb 7, 2021 at 16:21 | history | edited | Ivan Solonenko | CC BY-SA 4.0 |
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| S Feb 7, 2021 at 16:15 | history | answered | Ivan Solonenko | CC BY-SA 4.0 | |
| S Feb 7, 2021 at 16:15 | history | made wiki | Post Made Community Wiki by Ivan Solonenko |