No @Ted. The professor's approach is to start with the Eilenberg-Steenrod axioms, prove stuff that must hold for every homology theory and then actually construct singular homology and show it is in fact an homology theory
for sure once I know that the fourth line intersects the quadric in 2 points, there is at least one line for each point that intersects all four lines, correct? So the problem is just proving that there's not more than two?
We defined reps/subreps/irreps, showed irreps of abelian groups are 1-D, then did Maschke's theorem and Schur's lemma, now we've started character theory
@Karl: As I suggested yesterday, you have to argue from the geometry of the quadric surface why a point will in fact give you a line intersecting the other three lines.
yeah, this is the point that escapes me. for every point in the quadric Q there is surely a line passing through it. And that line will be one that intersects my three initial lines, by construction I would say
Ah, I see. Yeah we all had at least two quarters of algebra before this
Yeah it's definitely not too bad. Character theory seems exciting, we haven't proven anything yet, just computed characters of the irreps of $\mathbb{Z}/3$ and $S_3$ and talked about character tables
He just kinda defined them as an invariant that he promises will be easier to compute than just doing the matrices. He did mention that the weighted rows of the character table are orthogonal, as are the columns, so that's starting to make said promise seem more plausible
So you're looking only at the family of lines that meet the other three lines. In other words, those original three lines will never be a line that you pick through a point. It's worth understanding the two families of lines, but I see your point.
Well, I just don't understand yours. I am looking the quadric Q that is obtained by considering the lines that intersect my three original lines. I know that there is a fourth line intersecting Q in two points. Why do I need anything but my argument
@Daminark The first part is done. The proof is not quite trivial, but the statement is fairly simple. When you proved Maschke, did you do so by turning a linear map into a homomorphism of representations?
@BalarkaSen I think the way to think about the Riemann-Hurwitz realization problem is probably in terms of representations of free groups in the fundamental group, where conjugacy classes of specified words map to elements of specified order.
the proof for the fact that the inner product of two characters gives the dimension of the Hom space is a good motivation to consider traces, $f \mapsto \frac{1}{|G|} \sum_{g \in G} gfg^{-1}$ is a projection $\mathrm{Hom}_k(V,W) \to \mathrm{Hom}_{k[G]}(V,W)$, so it's trace is the dimension of the subspace, then you use the isomorphism $\mathrm{Hom}_k(V,W) \cong V^* \otimes W$ and use properties of traces
I think the number of cycles in the permutation corresponds to the number of circles above your given circle, while the length of the cycles corresponds to degree.
Ah yeah I guess that's a good way to say it. At the time we hadn't quite defined a homomorphism of $G$-reps so it wasn't stated in such terms, but that does make sense
@Daminark So if you do this to a linear map $f$ from an irrep $V$ to itself, you get a scalar. And that scalar is precisely $\frac{1}{\dim(V)}\operatorname{Tr}(f)$.
@manooooh: You have a quantity (represented by a letter, say $a$) that can take on different values. When you say it is fixed, you are setting it equal to one of those possible values. Example: if $a$ can be any integer from $1$ to $6$, I can say: "Fix $a=4$. Then I consider what happens only when $a=4$.
Well, you have to go back and see what the two families are (as I said about 10 minutes ago up there ...) and verify that your original three are all in one of those. @Karl
@MatheinBoulomenos The formula for the scalar I mentioned above actually turned out to be more tricky to prove than I expected. Not very tricky, but not any easier than the general version for symmetric algebras (I had hoped it would be simpler for group algebras, but not that I could make it)
@MatheinBoulomenos Yeah, though the scalar need not always be as nice. The main statement is that as long as the irrep is split simple, it will be some multiple of the trace (and the multiple depends on the irrep, not on the linear map)
This multiple is called the Schur element of the irrep (I think I have mentioned this before)
@manooooh But what that means is, "take one of these arbitrary constant numbers, any one of them, and then make $k$ itself constant and equal to that value"
@manooooh: Here's an example. I can consider circles of various radii centered at the origin. They're all of the form $x^2+y^2=k$ for $k>0$ a constant. But when I change circles, I change the value of $k$.
Because in some contexts it might be free to vary during the work at hand. We say "fix" when we want to be totally clear that the value cannot change once you start.
@LeakyNun First, we note that actually we only need to understand $SL_2$ since moving up to $GL_2$ is just a matter of twisting by some power of the determinant rep
@LeakyNun Next, we note that the diagonal matrices give us a maximal torus $T$, and that any irrep of $T$ is specified by sending the diagonal matrix $(a,a^{-1})$ to $a^n$ for some integer $n$
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@LeakyNun So we can denote any such irrep simply by that integer $n$, and we can inflate this to an irrep for the lower triangular matrices $B$ by having the lower triangular unipotent matrices act trivially. We still denote this irrep by $n$
Okay, even if we say "fix $k> 0$ an arbitrary constant" without mentioning any concrete value then $k$ has a value? Does it have a hidden value, right?
@LeakyNun Now we induce $n$ from $B$ to $G$, and a few magical things happen (we induce inside the category of algebraic representations). First, the result is $0$ unless $n\geq 0$. And if $n\geq 0$ we get a finite dimensional representation
1. representations of GL2 can be made from twisting representations of SL2 by some power of the determinant rep 2. the diagonal matrices is a maximal torus T and irreps of T correspond to integers 3. irreps of T give rise to irreps of B the lower triangular matrices 4. we "induce n from B to G"
But we can actually be more explicit here. $SL_2$ has an obvious action on the space of polynomials in $2$ varables (by letting those variables be the basis vectors in the natural representation)
In general, if $V$ is a rep of $H$ which is a subgroup of $G$, we define the induced representation by $\{f: G\to V\mid f(gh) = h^{-1}f(g)\}$ for $g\in G$ and $h\in H$
@TedShifrin I find a family of lines parametrised by $(a,a,b,b)$ and one parametrised by $(c,d,c,d)$. My initial lines where parametrised by $(a,b,0,0)$, $(0,0,c,d)$ and $(a,b,a,b)$. On the right track?
@LeakyNun $gh$ is an element of $G$, so $f(gh)$ is an element of $V$. And $f(g)$ is an element of $V$, so $h^{-1}f(g)$ is also, since $h\in H$ and $H$ acts on $V$.
it is more or less what it would mean for $f$ to be a homomorphism of representations of $H$, except acting on the right in each case (and of course with the action on $G$ not being linear)
It might be better to just ignore the technical stuff and just take the explicit construction instead. So we get a nice rep for each non-negative integer by acting on homogeneous polynomials
Which of the elements of $\mathbb{Z}_n$ are nilpotent? The only answer I could come up with is that $\bar{x} \in \mathbb{Z}_n$ is nilpotent iff $x^m \equiv 0 (mod \ n)$ for some $m \in \mathbb{N}$
@ÍgjøgnumMeg Hmm I'm not sure, I'm guessing that we'd only get nilpotent elements in $\mathbb{Z}_n$ if $n = p^k$ for some prime $p$, because like for example $\mathbb{Z}_6$ (seems) to contain no nilpotent elements apart from $\bar{0}$
Note that this looks a bit like the natural rep for $SL_2$, except everything has been raised to the $p$'th power
And in fact, raising all entries to the $p$'th power is of course a homomorphism from $SL_2$ to itself, so what we have done here is twist the natural rep by this homomorphism.
A useful thing to do in this sort of setup is to look for a vector which is fixed by the upper triangular matrices with $1$'s on the diagonal, and then check how $T$ acts on that vector.
Now, we can twist any representation that that homomorphism which raises all entries to the $p$'th power. If we do this to a representation $V$, let us call the result $V^{(1)}$ (we have twisted $1$ time).
