Timeline for Infinite dimensional representations of $\frak{sl}_2$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 14, 2023 at 1:15 | answer | added | Jeffrey Adams | timeline score: 4 | |
| Mar 13, 2023 at 18:09 | comment | added | LSpice | @მამუკაჯიბლაძე, re$\DeclareMathOperator\SL{SL}$, a cyclic representation need not be irreducible. In a representation of the algebraic group $\SL_2$, a la Jantzen, it will always be finite dimensional. I now see that there are representations of $\SL_2(\mathbb C)$ that are not algebraic in this sense; you have given one such. But your original representation was instead of $\mathfrak{sl}_2(\mathbb C)$. | |
| Mar 13, 2023 at 17:55 | comment | added | მამუკა ჯიბლაძე | @LSpice Please understand me correctly too, I am not countering any of your statements, I just want to understand them better. I still do not understand why I cannot take some infinite-dimensional representation of $\mathrm{SL}_2$ and generate a subrepresentation by a single vector. I believe such subrepresentations will always be irreducible and almost always infinite-dimensional, no? | |
| Mar 13, 2023 at 17:36 | comment | added | LSpice | @მამუკაჯიბლაძე, re, a consequence is that there is no representation of $\operatorname{SL}_2$ on your original representation that is compatible with the action of $\mathfrak{sl}_2$. In my comment, I was not trying to critique anyone else's reasoning, only to point out a flaw in my own. | |
| Mar 13, 2023 at 16:37 | comment | added | მამუკა ჯიბლაძე | I confirm that eigenvalues of $[x\partial_y,y\partial_x]=x\partial_x-y\partial_y$ on this representation are $-1,-3,-5,...$ The Weyl group, which in this case can be represented as switching $x$ and $y$, interchanges these with weights of another irrep spanned by $x^{k-1}/y^k$, $k=0,1,2,...$ But what is the consequence? | |
| Mar 13, 2023 at 16:11 | comment | added | LSpice | @მამუკაჯიბლაძე, re, I meant only to change the "actor" from the Lie algebra to the group, not the space being acted on. If I'm computing correctly, your original representation has as weights precisely the odd integers $\le -1$. If $\operatorname{SL}_2$ acted on the same space, compatibly with the action of $\mathfrak{sl}_2$, then $\left(\begin{smallmatrix}&1\\-1\end{smallmatrix}\right)\in\operatorname{SL}_2(\mathbb C)$ would convert each ($-(2k-1)$)-weight space to a ($2k-1$)-weight space. | |
| Mar 13, 2023 at 7:03 | comment | added | მამუკა ჯიბლაძე | @LSpice I tried to integrate my example above, what I get is the space of rational functions of the form $r(x,y)=1/(\lambda_1x+\mu_1y)+...+1/(\lambda_kx+\mu_ky)$. Unless I am overlooking something, this is an irreducible $\mathrm{SL}(2)$-subrepresentation of the space of all functions. Could you explain your remark about the Weyl group? | |
| Mar 12, 2023 at 15:30 | history | edited | LSpice | CC BY-SA 4.0 |
Deleted the reference to my wrong comment, *pour encourager les autres*
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| Mar 12, 2023 at 15:30 | comment | added | LSpice | @EBz, hmm, as foretold, I embarrassed myself. I had in mind the analogies: representations of complex groups and Lie algebras might as well be algebraic; and representations of algebraic groups are locally finite, so irreducible ones are finite dimensional. I guess the (a?) wrong step in my reasoning is that the representations of the Lie algebra that you describe don't integrate to the group (and, in particular, their weights need not be preserved by the Weyl group), right? | |
| Mar 12, 2023 at 15:17 | comment | added | მამუკა ჯიბლაძე | Yes. For a specific example, consider the representation on rational functions in two variables, as the Lie algebra generated by $x\partial/\partial y$ and $y\partial/\partial x$. The submodule generated by, say, $1/x$ is irreducible and infinitedimensional: it is spanned by $y^{k-1}/x^k$, $k=1,2,3,...$. | |
| Mar 12, 2023 at 15:11 | comment | added | user108998 | @ L Spice maybe I'm misunderstanding you but it is not the case that the irreps are finite dimensional. Indeed a Verma corresponding to non- (integral dominant) is irreducible isn't it? | |
| Mar 12, 2023 at 14:45 | history | edited | László Szabados | CC BY-SA 4.0 |
added 170 characters in body
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| Mar 12, 2023 at 14:19 | comment | added | Balazs | One possible answer is the Beilinson-Bernstein localisation theorem. Others much more knowledgable than me will weigh in here but let me just point to arxiv.org/pdf/2002.01540.pdf which works out some very concrete examples for ${\frak sl}_2$ (both complex and real). | |
| Mar 12, 2023 at 14:13 | comment | added | LSpice | I must be missing the meaning of your question in some embarrassing-to-me way. $\mathfrak{sl}_2$ means $\mathfrak{sl}_2(\mathbb C)$, right? Then isn't it true that its irreducible representations are finite dimensional, the representation generated by a highest-weight vector $v$ with weight $\lambda$ is spanned by a finite number of $F^i v$ (since $F^i v = 0$ once $w_0(\lambda - i\alpha)$ is greater than $\lambda$), hence is finite dimensional, and all representations are direct sums of irreducibles? | |
| Mar 12, 2023 at 13:39 | history | asked | László Szabados | CC BY-SA 4.0 |