@AlessandroCodenotti what's a function that's left-Haar-square-integrable but not right-?

Ugh @Leaky. You're gonna need a non-compact group, of course.

How do I show that closure of the set of matrices of rank $k$ is actually the set of matrices of rank at most $k$ for Zariski topology on the set of $m$ by $n$ matrices? I only have shown that matrices of rank at most $k$ are closed? Any ideas on how to proceed?

@TedShifrin right

@user330477: We already talked about this idea, actually. In the case of rank $1$ for $2\times 2$ matrices, do you see how to do it?

How do you find a Zariski-open set containing a matrix of rank $2$ and not the matrices of rank $\le 1$?

Hmm, test. Having difficulty posting from mobile, it seems

Nope, that went though.

@Rithaniel: Yesterday I was having constant trouble and gave up.

@TedShifrin Matrices of rank $2$ are non-invertible and thus open. But, when did we talk about this idea?

@TedShifrin so the derivative of your trouble is zero?

Maybe what I'm trying to post is too long. Is a question about path connectedness implying connectedness. I was walking through my attempt at a proof and pondering what I'm missing.

@user330477 I think matrices of rank $2$ are invertible

When I asked you what it meant to find the closure.

That was a mental slip-up, @Leaky, but good catch.

@user330477: But how do you distinguish a matrix of rank $>k$ from matrices of rank $\le k$?

interesting question

@TedShifrin I am not sure. But if a matrix has rank $k$, then there is a maximum of $k$ linearly independent columns.

Yeah, that won't give equations. You want equations.

@TedShifrin can I use the (removed) thingy?

@Leaky: It's equivalent to what I had in mind. You also need to give functions/equations.

right?

I was texting someone and missed it.

@LeakyNun Do what $k$ at time to get equations?

A bit vague.

LOL, right.

Give a hint, Leaky.

I don't know how to give a hint without giving away the answer lol

@user330477: How do you tell with an equation that $n$ vectors in $F^n$ are linearly dependent?

Oh, bad letter.

aha

I have another trick :P

better than wedging k columns at a time

No, that's not gonna do it, @Leaky.

it involves making a rectangle into a square @TedShifrin

A square contained in the rectangle? Yes.

right

That's where I'm trying to head @user330477.

but it's not a retract... the way you word it, lol

I'm trying to find a non-Hausdorff space $X$ such that every point in the space has a neighborhood homeomorphic to $(0,1)$. I tried the cofinite topology and particular point topology on $\Bbb{R}$, but these don't seem to work. I could use a hint.

Huh? It's a deformation retract.

Oh, this is the famous example, @user193319.

I mean, I don't see how I do it as retracting to that particular square

@TedShifrin I have a feeling this has got something to do minors of a matrix? Is there something more elementary?

No, that's it, @user330477. You need to think about precisely those.

P.S. That's totally elementary.

@user193319 oh right, that famous example

P.S. not very elementary for me

Dang it...

What's this famous example?

If I keep it up, I can infuriate everyone in the room.

@TedShifrin should I tell him lol

I'll give you a name, @user193319, and you figure it out.

I can't think of anymore non-Hausdorff space.

I called it the real line with two origins.

right

That's all I'm saying, @Leaky. :)

Oh, interesting. Thank you.

I already infuriated Mike, who says he'll never come back, because I wasn't using my room-owner prerogative to shut people up.

Eventually, I may do so, but I'm still waiting for you guys to draft rules we should agree on.

@user193319: You can write it down explicitly as a quotient space. If you want.

It's a good way to understand when the quotient topology fails to be Hausdorff.

@TedShifrin why won't wedging do the trick?

@TedShifrin Is it true that an $m$ by $n$ matrix has rank less than $k$ iff every $k$ by $k$ minor of that matrix is $0$?

Wedging the full vectors? Or sub-vectors?

yes, @user330477.

Think about the contrapositive.

wedging $k$ columns together

full vector

So that maps you to a huge-dimensional space, @Leaky. Then what?

then if every coefficient is zero then the original columns are linearly dependent

@TedShifrin oh and I think we have different approaches then

It turns out to be equivalent approaches, I think.

well...

Exterior algebra is just determinants; it can't be different.

@TedShifrin I see how to prove the forward direction, but not the backward one. Regardless, I am not sure how this would help in the actual problem, though?

communicating in codes is fun

I don't mean exterior algebra

I mean my approach without exterior algebra is different from your approach

Do the contrapositive of your other direction, @user330477.

so, the inverse, lol

@TedShifrin If $M_k$ denotes varieties of rank at most $k$, then after doing all this I only need to show that closure of $M_{k} \setminus M_{k-1}=M_k$, right? How would I do that?

@TedShifrin so it's safe for me to shout my answer right

It's equivalent to what we're already talking about, isn't it, @user330477?

Go ahead, Leaky. I'm not a tzar here.

so i'm looking at $\det(A^\top A)$

@TedShifrin I don't know what an exterior algebra is? The course I am taking is just a bit of algebraic geometry at the undergraduate level.

I hope I'm not horribly wrong lol

Ignore that, @user330477.

If $BA^\top A = I$ then for $Ax=0$ we have $BA^\top Ax = 0$ so $x=0$

Where $A$ now has $k$ columns in $F^n$, Leaky? So that square matrix is nonsingular iff $A$ has rank $k$. That's a good linear algebra fact, yes.

right

does that count as a different approach? :D

But you've got to do that for $\binom nk$ sets of columns?

right