@TedShifrin how do you like that definition? Let $M$ be a smooth manifold. Then a vector bundle on $M$ is a continuous and cocontinuous endofunctor of the category of $C^\infty(M)$-modules. A morphism of vector bundles is a natural transformation between such functors
@Ted is it just that you felt faculty were favoring grad students a bit too much at you wanted to give extra attention? Or is there something about teaching undergrads that you prefer?
Demonark: A lot of students came into my Calc Theory (Spivak) and Honors Multivariable (my book) courses having no intention to pursue math and ended up math majors.
@TedShifrin I wonder if that is a cultural thing? Here, it is hard to get the undergrads to get excited about much of anything (a surprisingly large number of the math undergrads are former physics majors who couldn't hack it in physics), but the grad students are generally very on top of things.
Interesting, Xander. I definitely got a lot of interest generated with undergrads. Some of the grad students avoided me (and their advisers even recommended that) because I demanded hard work to get an A.
Hmm, I've noticed in analysis that the grad students weren't quite as motivated, though I chalked it up to the fact that the department encourages grad students interested in analysis to place out of things, and this year in particular we didn't have many grad students interested in analysis
I feel like in undergrad courses you're just one of many students, but when you're in a grad class with 4 other people, the profs really get to know you
As an adviser, I tried to suggest the best teachers/courses depending on my students. I did make my one PhD student take a functional analysis class for her own good, even though the teacher was not optimal.
@Mathein: My undergrad students got to know me as much as they wanted to ... some just didn't.
I was a very different sort of teacher from the European model.
I think there's a sense in which grad students may want to kind of zoom in more on the specific things they're interested in? So maybe they feel like engaging with other subjects is inconvenient somehow, or at least a suboptimal way of spending time
(on prof who's a bit unconventional talked to me about l-adic cohomology and Galois representatiosn when I was a Freshman, I didn't understand anything)
I think being broad is kinda better. It's probably not my place to say much given that I'm talking about a context that I'm not even in yet, but part of it is that there are so many interesting areas and it's sorta cool to just see some magic come outta throwing seemingly disjoint stuff together
Demonark: One of my students in grad diff geo persisted in the course, over the objections of his adviser, because he knew that geometry would be relevant in number theory (hyperbolic geometry, in particular). He was great.
Also I remember Neves mentioning that he made good use in knowing just enough about other areas to make analogies with his own work, so it seems useful to have at least a passing acquaintance with ideas even if they're not interesting to you in themselves
breadth is nice and great, but there's also a lot of breadth in depth if that makes sense. I mean most fields themselves are huge with so many subfields etc.
@Mathein that fact always surprises me. Like, if you naively think about what you expect number theory to connect with, you'd expect that it's somewhat more narrow than other things. Like, I've heard that ergodic theory of all things has application to number theory
Regarding applied stuff, I can sorta buy a case that it's healthy to see a field in context from a general academic perspective, but that if an exchange is sufficiently one-sided, there's a case that perhaps making it essential is excessive
Ah, I thought it was more standard to do that. In that case I'd probably try to have a major which doesn't place too many specific requirements so people can mold it as they please
that's exactly what you show in the general proof of Maschke's theorem
if you know that invariant subspaces have invariant complements, just take the subspace spanned by an eigenvector, an invariant complement of that, restrict to the complement, repeat
@LeakyNun I'm a bit surprised how many people learn LA without doing minimal polynomials
you can define what an algebraic structure is in universal algebra and you can define what an algebraic category is in category theory and fields are neither of that
@MatheinBoulomenos: I'd like to know how to properly formulate (mathematically) a concept called "descent in symmetry".
@MatheinBoulomenos: You have a group $G$ and "descent in symmetry" with respect to an iirep of the group, by checking which characters (now lets only use 1-dim irreps (over $\Bbb R$)) are not 1. These operations you remove from the group and you get a subgroup. Thats it.
@MatheinBoulomenos: Could be since they call it the "kernel of $G$ with respect to irrep", for higher-dim irreps you get also what they call "epikernels".
@Rudi_Birnbaum the kernle of a one-dimensional representation is just the set of elements with character $1$, that's a subgroup. But the set of elements with character not $1$ is not a subgroup
While the kernels form so to say a series of "subgroups" the "epikernels" are all subgroups of $G$ but there are more than one and they do not include each others strictly.
@MatheinBoulomenos: Sounds great! What now with higher dim-irreps?
@Rudi_Birnbaum I googled "epikernel" and apparently, for each vector of the space on which the irrep acts, you look at the subgroup that fixes this vector, that's a subgroup (we would call that a stabilizer subgroup)
okay, this implies that it's the intersection of all epikernels, or in other words, it just consits of all vectors that act like the identity for this irrep
@MatheinBoulomenos: Sorry to bother you, but could you probably make a definition of that? Like: "A descent in symmetry from a group $G$ along an irrep $\Gamma$ is ...
I'd formulate it like this: for a representation $\rho: G \times V \to V$, the kernel is the set of all $g \in G$ such that for all $v \in V$ $\rho(g)v=v$ For every fixed $v \in V$, we have an associated epikernel which is the set of all $g \in G$ such that $\rho(g)v=v$
@Rudi_Birnbaum that's something you can prove! it follows from the properties of representations. If $\rho(g)v$ and $\rho(h)v$, then $\rho(gh)v=\rho(g)\rho(h)v=\rho(g)v=v$ etc.
@MatheinBoulomenos: And I miss somehow the term "descent in symmetry", it is very important, that this appears, since I need this to communicate with mathematicians.
If $G$ is a point group, then descent in symmetry is the map that maps the pair $(G,\rho)$ to the pair consisting of the kernel and the set of epikernels. (note:the kernel can be recovered from the epikernels by taking the intersection over all of them)
which is basically what you said
if you want to communicate with mathematicians, then the term "stabilizer subgroup" for the epikernels will be helpful
you can say: for a representation $\rho: G \times V \to V$ and a vector $v \in V$, the epikernel associated to that vector is the stabilizer of that vector
@MatheinBoulomenos: Perfect, thanks a lot! (Just one more minor thing: Is there a way (map, name of it?) how to trace where the irreps from the group ended in the epi(kernels)?
@MatheinBoulomenos: Or do the get completely "shaken up"?
@Rudi_Birnbaum when you have a group $G$ and a subgroup $H \leq G$, then you can restrict the representation of $G$ to $H$ (and you can decompose that into irreducibles of $H$)
@MatheinBoulomenos: I need to understand if there is more information if the group subgroup relation comes from a descent in symmetry than if its "just" a group-subgroup relation. Or maybe the other way round can I get from the group to all subgroups by the descent in symmetry?
but I think if you allow all representations for a group, then you can get every normal subgroup as a kernel and every subgroup as an epikernel, so it might be more interesting to just look at irreps
@MatheinBoulomenos: That makes absolutely sense (from point of view of my application)
@MatheinBoulomenos: You see the iireps are vibrations of a molecule.
@MatheinBoulomenos: And some molecules in high symmetry are unstable with respect to some vibrations. So they descend in symmetry in that "direction". But that only happens in propper "vibrational modes". Just as a side note.
@MatheinBoulomenos: So again, how to define the map from the irreps of $G$ to those of the stabilizer group.
that's something you would have to spell out because it's not something that exists commonly in rep theory, afaik. Here's the technical definition, as I understand it: Consider the set of all epikernels of a group $G$ (obtained from possibly different irrep), then we say that for two epikernel $K_1$ and $K_2$ that $K_1$ is "contained in symmetry in $K_2$ if $K_1 \subset K_2$ and there exists an irrep $\rho$ such that both $K_1$ and $K_2$ are epikernels with respect to $\rho$
@Rudi_Birnbaum the thing is, when you restrict an irrep of $G$ to a subgroup $H$, it might not be irreducible anymore. and typically, you're not just interested which irreps of $H$ occur in that resctriction, but also with which multiplicity
of course all this is information is contained in the character: you just restrict the character of $G$ to a subgroup, that's the character for the restricted representation
@MatheinBoulomenos: These calculation details are not so interesting, I know how to do that anyway. For me its much more important how to formulate things properly. I'll need that since I have a conjecture.
@Rudi_Birnbaum I don't think there's a name for that. you would just write out something like "consider the decomposition of the restriction to $H$ into irreducibles" or work with the character since that contains all the information
@MatheinBoulomenos: "Consider the set of all epikernels of a group $G$ (obtained from possibly different irrep), then we say that for two epikernel $K_1$ and $K_2$ that $K_1$ is "contained in symmetry in $K_2$ if $K_1 \subset K_2$ and there exists an irrep $\rho$ such that both $K_1$ and $K_2$ are epikernels with respect to $\rho$": I am not sure if that is really what I want, because I miss here the reference to the other irreps not those along which we (distort=) descend in symmetry.
@MatheinBoulomenos: Do we agree, that the map I search for is the map from the "remaining" (all others except $\rho$) irreps to the irreps from the epikernels?
@MatheinBoulomenos: I understand from my applications that there some ambiguity, but also some information retained about which $\Lambda$ has to be associated with which $\Lambda$ from the stabilisers.
@MatheinBoulomenos: $G$ has a set of irreps, how to call them?
@Rudi_Birnbaum so if one subgroup is obtained as epikernels for two different irreps, do you want to have two distinct copies or do you still consider it as the same? You'd have to decide that
So we have ($G,\{\rho_i\}$) and descend in symmetry along $\rho_k$ to get say in this case only one ($K1,\{\sigma_j\}$). Now I want to know if there is generic map from the set \{\rho_i\}/\{\rho_k\} to \{\sigma_j\}
@MatheinBoulomenos: A simple example would be that one part of this map is always the trivial representation of $G$ to the trivial representation in $K_1$.
@MatheinBoulomenos: I doubt this. But could that be in case it exists a peculiarity of point groups?
Suppose you and someone on social media both like to follow the local teams: basketball, baseball, and football. (Note that I"m pretending you're American for the purpose of this :P)
Each night, each of you choose to watch one of the games and report on social media whether the home team one or not. If you both watch the same game, then you'll both of course report the same outcome (which I'll call + or -)
@MatheinBoulomenos: So the point is that sometimes a higher dimensional irrep splits into several new irreps in the stabilizer. And I am now searching for a functional relation between functions of the two.
So if you were to keep track of your various social media accounts, you'd find that P(first person saw the home team win, second person saw the home team win | first person watched hockey, second person watched hockey) = 1
@MatheinBoulomenos: say $\lambda_{\rho_k}(\rho_j) = (\sigma_a\oplus\sigma_b)$. Then I have conjecture connecting $\rho_k \otimes \rho_k$ and $\sigma_a\otimes\sigma_b$
I got into a spat with my learning partner today, he wanted to write the problem statement using the standard basis but I claimed you didn't gain any insight or anything useful for the context of the problem by fixing a basis at all
feel free to think of it like that: each of you get to watch one of three coin flips.
if you watch the same one, you get the same outcome
and so forth
anyways. since the partner's choice of which game they watch shouldn't matter as to whether the coin is fair, that requirement further imposes that P(+|H) = P(++|HB) + P(+-|HB) = 1/2
