Timeline for New (?) math object. Looking for (if existing) literature [closed]
Current License: CC BY-SA 4.0
27 events
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| Apr 30, 2022 at 14:24 | history | closed |
LSpice YCor Steven Landsburg skupers Neil Strickland |
Needs details or clarity | |
| Apr 30, 2022 at 12:24 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
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| Apr 30, 2022 at 8:24 | comment | added | YCor | Could anybody who understood the OP's intent edit the question so as to make correctly stated? | |
| Apr 30, 2022 at 7:32 | comment | added | Geoff Robinson | Notice that if $H$ has that property, so does its linear span. After this extension, you are basically looking at $G$-invariant subspaces of ${\rm End}(V)$. | |
| Apr 30, 2022 at 6:28 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Apr 29, 2022 at 23:15 | comment | added | Fabio | Sorry I am not a mathematician but the way sam expressed my question is the way I do understand the problem. Is this something that people have studied? If not what are similar proprieties that I can start to look at that people have already studied? A group ‘normalized’ by a set of operators? Thx | |
| Apr 29, 2022 at 23:12 | comment | added | Andreas Blass | As @LSpice said, the sentence "Define $H\ \dots$ in the question doesn't define $H$. The requirement $(\forall g\in G)\,gHg^{-1}=H$ is satisfied iff $H$ is the set of all operators on $V$. It's also satisfied if $H=G$. It's also satisfied if $H$ consists only of the identity operator. It's also satisfied if $H$ is empty.The "definition" fails to say unambiguously what $H$ is. | |
| Apr 29, 2022 at 23:09 | comment | added | Matt Zaremsky | I would call this, "$H$ is normalized by $G$," but I think the thing causing the debates is that you can't say "the" group $H \le GL(V)$ normalized by $G$, but rather "a" group $H \le GL(V)$ normalized by $G$. For example $H=\{1\}$ is normalized by $G$, and $H=G$ is normalized by $G$, and $H=GL(V)$ is normalized by $G$, and probably various other choices of $H \le GL(V)$. So it's not like a given $G$ leads to a uniquely determined $H$. | |
| Apr 29, 2022 at 22:30 | comment | added | LSpice | @SamHopkins, re, if the asker confirms that this is what they meant, then I am not here just to harp on such issues of language. However, since at least one user (other than I) was confused, it seems worth it to be explicit; and the asker has not yet explicitly agreed to this interpretation, so I'm still not sure that it's what they meant (though I agree that it seems plausible, now that you've suggested it). | |
| Apr 29, 2022 at 22:27 | comment | added | Sam Hopkins | @LSpice: I think this is just a language thing. It would be normal to say, informally, "Let $X$ be a set where you can add its elements and scale them; what is a name for $X$?" with the intended answer answer being "vector space." | |
| Apr 29, 2022 at 22:23 | comment | added | Fabio | Thanks Lspice and Sam. Lspice do you have in mind any pointer to help me getting an idea of H as a random a H⊆End(V)? I have the feeling that even for simple groups it a difficult object to characterize. | |
| Apr 29, 2022 at 22:20 | review | Close votes | |||
| Apr 30, 2022 at 14:24 | |||||
| Apr 29, 2022 at 22:19 | comment | added | LSpice | The definition is circular because, under the requirement $\forall h \in H\,\forall g \in G\,\exists h' \in H\,g h g^{-1} = h'$, you cannot look at an individual operator $h$ and tell whether or not it belongs to $H$; you need the whole family $H$ first. You could, as @paulgarrett suggests, require that $h' = h$; or you could, as @SamHopkins suggests, ask of a random $H \subseteq \operatorname{End}(V)$ whether it satisfies this property—but you cannot so define $H$. | |
| Apr 29, 2022 at 22:15 | comment | added | Sam Hopkins | @Fabio: sorry, can't help you with that. It's a natural thing to consider but I've never heard of any special name. | |
| Apr 29, 2022 at 22:06 | comment | added | Fabio | Yes Sam! thanks and some pointer in the literature. | |
| Apr 29, 2022 at 22:04 | comment | added | Fabio | Let me try again: H is a set the of operators that are mapped into themselves (modulo permutation) by G-conjugation (G given). Why is the definition of H circular? | |
| Apr 29, 2022 at 22:02 | comment | added | Sam Hopkins | I think Fabio is asking for a name for the property of a subset $H$ of linear endomorphism of $V$ being closed under conjugation by $G$. | |
| Apr 29, 2022 at 21:59 | comment | added | LSpice | That condition is circular: you can't test whether $g h g^{-1}$ belongs to $H$ until you've defined $H$, so you can't use that requirement in the definition of $H$. | |
| Apr 29, 2022 at 21:31 | comment | added | Fabio | I mean the set of operators $H$ with the property that for all $h\in H$ $ghg^{-1}=h',\;\;\exists h'\in H\;\; \forall g\in G$. | |
| Apr 29, 2022 at 21:28 | comment | added | Fabio | Thanks LSpice, I corrected it, typo | |
| Apr 29, 2022 at 21:27 | comment | added | LSpice | "Define $H$ as the set of operators $H = \{h : V \to V, h \in H\}$ …"? I'm having trouble decoding that definition, which sounds circular. In $g H g^{-1} = H,\quad\forall g \in G$, do you mean $g h g^{-1} = h,\quad\forall g \in G, h \in H$? | |
| Apr 29, 2022 at 21:27 | history | edited | Fabio | CC BY-SA 4.0 |
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| Apr 29, 2022 at 21:26 | comment | added | Fabio | Thanks paul, can you point me out a easy paper on the topic? | |
| Apr 29, 2022 at 21:24 | comment | added | Fabio | thanks carlo, why is that? what about other groups like SU(n) for example? | |
| Apr 29, 2022 at 21:22 | comment | added | paul garrett | It'd be called the "commutator" of (the image of) $G$ in operators on $V$. Yes, much-studied... | |
| Apr 29, 2022 at 20:50 | comment | added | Carlo Beenakker | do I understand this correctly? take as an example for $G$ the group of orthogonal matrices, then the operators $H$ are multiplication by a scalar. | |
| Apr 29, 2022 at 20:42 | history | asked | Fabio | CC BY-SA 4.0 |