Mathematics

12:41 PM

@TobiasKildetoft the highest weight of the usual $\mathbb C^n$ module is just $\lambda = \epsilon_1$, right?

@anakhronizein yeah

And for the dual representation, is it $-\epsilon_n$?

Well all I know is that $\dim (V^*)_\lambda = \dim V_{-\lambda}$ so I am actually not too sure on that one.

well, what is the highest weight vector in the dual?

Well the weight spaces look like $V_\lambda = \{f\in V^*\mid \forall v\in V\,.\,f(e_{i,i}v) = a_i f(v)\}$ where $\lambda = \sum a_i\epsilon_i$...

I thought it was the nth linear functional in the dual basis.

But I think that is wrong.

So you want an element that is killed by all the strictly upper triangular matrices

12:49 PM

killed with respect to the dual representation action, right?

right

Which I guess just boils down to finding which 1xn matrix kills an upper triangular matrix?

Which would have to be [1,0,...,0]?

Or some multiple.

Oh, no nevermind

So you want a linear map whose kernel contains the image of the strictly upper triangular matrices

[0,...,0,1]

This one works, doesn't it? Because if $A$ is upper triangular, then the columns of $A$ always have 0's in the last entry.

What does that notation mean? We are looking at linear maps now

12:53 PM

Sorry, I am thinking as matrices.

So if $\{\bar{e_1},\dotsc,\bar{e_n}\}$ is the basis for $\mathbb C^n$, then we take the $f_n$ in the dual basis such that $f_n(\bar{e_n}) = 1$.

ok

Is that right? It does kill them.

Right

So is this a weight vector, and what is the weight in that case?

So now the weight which corresponds to it is $\lambda = \sum_i a_i\epsilon_i$ where $f\cdot e_{i,i} = a_i f$.

Where $f$ is this $f_n$

But that's 0 unless $i=n$

Why are you acting on the right?

1:00 PM

Sorry, my bad, I meant matrix multiplication by $\cdot.$

So $f$ as an $1\times n$ matrix, $e_{i,i}$ as the $n\times n$

Yeah, I think that is a bad idea for this.

It yields the wrong answer??

So the highest weight is not $\lambda = \epsilon_n$...

probably not, but it makes you not think about duals in the "correct" way

How should I be looking at it?

Well, $f$ is a linear map on the space, and we know how the algebra acts on it

1:02 PM

By transpose?

Well negative.

right. It is better to not think of everything as matrices, since this whole thing really works in the abstract

So we have $(e_{i,i}\cdot f)(v) = -f(e_{i,i}\cdot v)$.

$e_{i,i}\bullet f_n := - f_n^T e_{i,i}$

I see.

Now apply this with $v$ being standard basis elements to see what it looks like

So $e_{i,i}\cdot \bar{e}_j = \delta_{ij}e_i$, right.

King Tut

Hello

1:06 PM

So $(e_{i,i}\cdot f)(\bar{e}_j) = - \delta_{ij}f(\bar{e}_i) = -\delta_{ij}$.

King Tut

if i need to study complex analysis, then what are the prerequisite

Real analysis? Calc 1,2,3? are there other topics

Ick, my first one broke.

And now I can't edit it, heh.

Shobhit

@KingTut IMO you'll only need elementry analysis, convergence, connectedness etc etc.

Oh well. I get $\lambda = \sum_j -\delta_{in}\epsilon_i = -\epsilon_n$.

King Tut

@shobhit ok thanks!

1:10 PM

Granted my TeX skills need work since now that I read what I wrote, it's riddled with msitakes.

But the conclusion (assuming you take my work modulo typographical mistakes) is correct, @TobiasKildetoft ?

@anakhronizein Yes, that looks right

Right, so as it relates to showing the standard C^n module is not necessarily self-dual, I am guessing there exists some theorem that says that isomorphic $gl_n$-modules have the same highest-weights (when they exist)?

yes

In fact, that is a very easy thing to show. The hard thing is to show that irreducible reps with the same highest weight are isomorphic

I assume it follows from the weight space decomposition, and then you look at the action of the representation on these spaces?

which direction?

The easy direction is just noting that an isomorphism sends highest weight vectors to highest weight vectors of the same highest weight

1:20 PM

@AkivaWeinberger well it's obvious if you use the origami

Oh yes you are right that is easy, @TobiasKildetoft

I will write this down to make sure I get everything. Thanks for your help!

The other direction is more involved. The way I know is by considering Verma modules.

Shobhit

Let $f \in M^{+}(X, \bf X)$, let $(E_{n})$ be a sequence of disjoint sets in $\bf X$ and define $$f_{n}= \sum_{k=1}^n f \chi_{E_{k}}$$ then i can see that $(f_{n})$ is an increasing sequence but how does it imply that $(f_{n})$ converges to $f \chi_{E}$, where $E = \cup_{n \in N} E_{n}$, where $f \in M^{+}(X, \bf X)$ means $f$ is a non-negative real valued $\bf X$- measurable function from $X$.

I have never heard of Verma modules, @TobiasKildetoft, is it worth looking into? Is there any good reference that touches on them?

@anakhronizein Humphreys covers all of this (though he considers semisimple algebras, the jump to the general linear one is not a major one)

1:30 PM

His intro to Lie algebras book?

Shobhit

got it, nvm

@anakhronizein Yeah

@TobiasKildetoft is it true that $H_n(X,A) = H_n(X/A)$?

I will have a look, thanks!

@LeakyNun In what sort of context?

1:39 PM

@TobiasKildetoft relative singular homology

no idea then

ok thanks

Looked into excision, @LeakyNun?

looking

When trying to find all left inverses of a matrix can one use the transpose to find all left inverses of the transpose matrix? For example, to find $X$ which satisfies $XA=I$

$$\begin{bmatrix}
x_{11}&x_{12}&x_{13}\\x_{21}&x_{22}&x_{23}
\end{bmatrix} \begin{bmatrix}
2&1\\-1&2\\3&1
\end{bmatrix} = \begin{bmatrix}
1&0\\0&1
\end{bmatrix}$$

Can I instead find $X^T$ which satisfies $A^T X^T = I$ (since $I^T=I$)

$$
\begin{bmatrix}
2&-1&3\\1&2&1
\end{bmatrix} \begin{bmatrix}
x_{11}&x_{21}\\x_{12}&x_{22}\\x_{13}&x_{23}

1:52 PM

@philmcole sure

I tried it and got

$$ X^T = \begin{bmatrix}
2/5&3/5\\-1/5&1/5\\0&0 \end{bmatrix} + t \begin{bmatrix} 9&0\\3&0\\-5&0\end{bmatrix} + s \begin{bmatrix} 0&9\\0&3\\0&-5\end{bmatrix}$$

but when I transpose it to get $X$ it seems to be wrong...

9 3 -5 should be wrong

Even if, setting $t,s=0$ is a valid solution right. But that doesnt work either.

Concretely,

$$ \begin{bmatrix} 2/5&-1/5&0\\3/5&1/5&0 \end{bmatrix} \begin{bmatrix}
2&1\\-1&2\\3&1
\end{bmatrix} \neq \begin{bmatrix}
1&0\\0&1
\end{bmatrix}$$

you forgot to transpose the first matrix

@philmcole also the first matrix here is wrong anyway

3/5 and 1/5 should be wrong

mmm ok I'll check

2:03 PM

so $D^n/\partial D^n \cong \partial D^{n+1}$

Tanuj

2:39 PM

@PrithiviRaj Hey man !