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20:01
I guess you're talking about $\det(I+AB) = \det(I+BA)$.
Yeah, there are matrix tricks for that, but I don't know a snappy exterior algebra argument.
Heya, @copper. Long time no see.
@TedShifrin Hi Ted, yep, busy getting ready for tomorrow.
Good luck having a new hip!!
Much appreciated.
@TedShifrin Let $\text{Hom}^k(E, F)$ be the set of $k$-linear symmetric maps from $E^k\to F$, and $Q^k(E, F)$ be the set of maps $g:E\to F$, $g(h) = f(h, h, ..., h)$ for some $f\in \text{Hom}^k(E, F)$. Then a formal power series consists of a sequence of maps $g_1, g_2, g_3, ...$ such that $g_k\in Q^k(E, F)$ and its convergent if $\sum g_n$ is convergent in a neighbourhood of $0$
I'm gonna have cataract surgeries in the new year, but I'm still jealous of the hips :)
20:10
This is a definition by Bochnak
To be honest, @Jakobian, I have studied several complex variables (in $\Bbb C^n$), but I don't know why I would be interested in infinite dimensions.
Good luck @Ted!
@copper I guess it gives you a good excuse to miss work and Christmas obligations :)
@TedShifrin Best of luck Ted.
@TedShifrin :-) To be honest, in some way I would be glad to defer...
20:14
Well, you'll love being able to walk more than a few blocks again :)
@TedShifrin Is it not? Take $v=(E_1,\dotsc,E_n)$, no?
The $I$ is coming from inside Koro's defined $E_i = (e_i,a_i)$. I'm confused.
@TedShifrin almost forgot that, but yeah, it has a convenient proof by checking it in the universal matrix ring where you can use that the universal matrix is invertible
Thanks, @Koro. I'll get the first stuff scheduled tomorrow.
We were trying to see some geometry — or at least I was. :)
ah yeah, sorry, $v$ is the gradient of $g$
where $g$ is the function whose graph's Gram matrix we're computing
20:18
$a_i= \frac{\partial g}{\partial x_i}$
in the notation $E_i= (e_i, a_i)$ for all $i\in [n]$.
Then column operation (last row -a_1* first row).
it gives -1-a_1^2 in the last entry.
etc.
I had to go earlier. I see Jakobian has extensively discussed this with you now and have not read your conservation, so apologies if I'm just regurgitating what has already been said, but here's how I would approach the problem:
If $R$ is a ring whose underlying additive group is finite cyclic of prime order $p$, pick an additive generator $x$. The elements of $R$ are then $kx$ for $k=0,\dotsc,p-1$. Thus, there is a constant $c\in\{0,\dotsc,p-1\}$ s.t. $x^2=cx$. This determines the multiplicative structure for $(kx)(lx)=klcx$. If $c=0$, the ring has trivial multiplication. Otherwise, $c$ is
Row operation, not column?
eventually one gets -1-a_1^2-a_2^2-... etc.
I used column.
(it is not necessary to assume $R$ commutative, that follows automatically)
20:22
minus can be taken care of by re-orienting.
I prefer my approach. I don't see how your column operations are computing the determinant of the matrix $BB^\top$.
So $A=[E_1, E_2,..., E_n, E_{n+1}]$, where , means a new line.
E_{n+1} is orthogonal to E_i for i in 1 to n and let's make it unit normal by dividing it by $\|,\|$.
Then I found det(A).
The first row of A is [1,0,0,..., a_1] and the last row is $[-a_1,-a_2,...,-a_n, 1]$. Applying column operation, these become resp. [1,0,...,0] and [$-a_1,-a_2,...,-a_n, 1+a_1^2]$. Then we expand the determinant along the first column.
and keep repeating it.
20:40
@Ted I'm stuck on something silly. Consider the Grassmannian of projective lines in $\mathbb{P}^n$. I have a proper subvariety $X$ of $\mathbb{P}^n$ (proper in the sense of containment) and want to consider the set of lines $L$ s.t. $L\cap X\neq\emptyset$. Is there an easy argument that the set of lines $L$ s.t. $L\subseteq X$ does not contain any component of the former? No use of the dimension of these spaces allowed, those are what I'm trying to compute.
21:06
actually, I'm not even sure if the set of $L$ s.t. $L\subseteq X$ is closed. if $L$ is the line going trough $p$ and $q$, I can express $L\subseteq X$ as a bihomogeneous condition in the projective coordinate of $p$ and $q$, but that doesn't necessarily translate to the Plücker coordinates of $L$...
21:35
@Thorgott What if this $X$ is a ruled variety?
I... will have to look up what that means
Union of linear spaces.
Like a quadric in $\Bbb P^3$?
Hyperquadrics are ruled by higher-dimensional linear spaces.
I don't see what goes wrong there?
For most varieties only finitely many lines are contained. Here there’s a large variety contained. That might be an embedded component or something.
the way I see it, for two points on the quadric, the condition that the line through them is contained in the quadric is given by a single bihomogeneous polynomial in those points coordinates, so the space of lines in a quadric should be $1$-dimensional still
the space of lines meeting the quadric on the other hand should have dimension $4$
21:52
Conway claims that the fundamental theorem of calculus is still true if we consider rectifiable paths (instead of piecewise smooth paths), at least for line integrals. I couldn't find a reference for this generalization.
this is his definition of a line integral (in terms of the Riemann-Stieltjes integral)
We doing the surface? Right. Two curves of lines contained (doubly ruled). Every line meets. But those curves are an embedded component or something in the whole Grassmannian.
I can of course have a look in the proof for the Riemann integral, and see if it translates
(which I'll do now)
my guess is now that we don't have the first fundamental thm of calculus, but we do have the second
oh right, every line meets (duh, that's even easier)
but then my first space is the whole Grassmannian and that only has one component
The Lebesgue integral has its own FTCs, Sha. Jakobian is an expert.
yeah, but there are these two special $P^1$s contained in it.
I'm not dealing with the Lebesgue integral though
I am familiar with the FTC for the Lebesgue integral
do they coincide in this case?
21:59
all I'm asking for is the $\{L\vert L\subseteq X\}$ does not contain a component of $\{L\vert L\cap X\neq\emptyset\}$, which is clearly the case if the latter is the entire Grassmannian and the former is not
@ShaVuklia I'm afraid your $\gamma$ need not be smooth?
I think the proof for the second FTC works just as well for the Riemann-Stieltjes integral
Oh, it is always a smaller-dimensional subvariety. Usually $0$.
so I consider my question solved
that sounds plausible, but I'm trying to use this fact to compute the dimension $\{L\vert L\cap X\neq\emptyset\}$
actually, though, how do you compute the dimension of $\{L\vert L\subseteq X\}$?
look at incidence correspondence, project both ways.
22:02
that's what I'm doing for $\{L\vert L\cap X\neq\emptyset\}$, but what incidence correspondence helps with $\{L\vert L\subseteq X\}$?
I might be missing an easy observation somewhere
@Thorgott correct. but that's not a problem here I believe (for my original question), since I just went through the proof of the FTC for Riemann integrals, and I could transfer it (mutatis mutandis)
@robjohn I'm thinking elf, or green mean square with a santa-cap (i.e., grinch). What up!?
well, my point is that if $\gamma$ were smooth, you would have $d\gamma(t)=\gamma^{\prime}(t)dt$ and could apply the actual FCT, but in general it's only a measure and you need some sort of FTC for the Riemann-Stieltjes integral, as you say
@Thor i think we want to look at linear fibers of the Gauss map of $X$.
Hi @amWhy
22:13
Gauss map in algebraic geometry?
Yes. The tangent plane must contain the line along the line. But it needn’t be the same, if course.
@TedShifrin Hello!
I guess we can work in affines and write down the polynomials defining $X$ and plug in parameterizations of lines, requiring they vanish for all $t$.
Happy holidays/vacation, @amWhy.
@TedShifrin Ditto, for you as well!
I’m a retired bum … all vacation :)
Better to think of chords of the variety, @Thor. Did you say that earlier? When are chords contained? That is an incidence variety set-up on $X\times X$.
22:25
@TedShifrin yeah, if $x,y$ spans my line $L$ and $f_1,\dotsc,f_m$ cut out $X$, then $L\subseteq X$ iff $f_i(sx+ty)$ vanishes identically as a polynomial in $s,t$ for $i=1,..,m$ iff the coefficients of $f_i(sx+ty)$ as a polynomial in $s,t$ vanish for $i=1,...,m$, but if I write those down explicitly, it seems these are bihomogeneous conditions on $x,y$ and don't only depend on the Plücker coordinate, which is odd
@TedShifrin How would you attempt to prove that second derivative is a symmetric bilinear map?
My usual proof is the classic MVT proof with continuity of the mixed partials. Dieudonné’s proof is much fancier and good for advanced audiences.
@Thor interesting. What if we try the explicit example of a quadric?
Sacrifice of simpler argument for increase in generality?
Some things are too sophisticated for first-year students. Dieudonné just assumes existence, not continuity, as I recall, of the derivative of the derivative.
If my quadric is $x_0x_3-x_1x_2$ and I take two points $[a_0\colon a_1\colon a_2\colon a_3],[b_0\colon b_1\colon b_2\colon b_3]$, I get that the line they span is contained in the quadric if and only if $b_0b_3-b_1b_2=0$, $a_0a_3-a_1a_2=0$ and $b_0a_3+a_0b_3-a_1b_2-b_1a_2=0$
the former two conditions just say that $a,b$ lie on the quadric, the last is the equality of two Plücker coordinates
so I guess it works out in this case
22:31
Yes, the function is twice differentiable but the second derivative doesn't have to be continuous. I never seen it stated like this
It’s elegant.
Gosh, I'm glad that course evaluations are not used at my institution with respect to pay / hiring / continuing contracts / etc.
I got one whole response this semseter.
but in general, if $X$ is cut out by $f_i=\sum_{j_0,\dotsc,j_n,\sum j_l=d_i}a_{j_0\dotsc j_n}^iT_0^{j_0}\cdot\dotsc\cdot T_n^{j_n}$ for $i=1,\dotsc,m$, the conditions on the line through $[x_0\colon\dotsc\colon x_n]$ and $[y_0\colon\dotsc\colon y_n]$ I get are that
$\sum_{j_0,\dotsc,j_n,\sum j_l=d_i}\sum_{k_0,\dotsc,k_n,\sum k_l=k}a_{j_0\dotsc j_n}^i{j_0\choose k_0}\cdot\dotsc\cdot{j_n\choose k_n}x_0^{k_0}\cdot\dotsc\cdot x_n^{k_n}y_0^{j_0-k_0}\cdot\dotsc\cdot y_n^{j_n-k_n}$ for $i=1,\dotsc,m$ and $k=0,\dotsc,d_i$, which is not quite enlightening
It is a very good response, but completely worthless. Need more $n$.
*ah, the indices should only run through $k_l\le j_l$
22:39
I almost feel like I've been lied to, since all this time I thought continuity of partial derivatives is pretty much required for Schwartz theorem to hold
the equations for $k=0$ and $k=d_i$ correspond to $[y]$ and $[x]$ resp. lying on $X$, but I don't see why the middle ones would be equations in the Plücker coords
@Xander Some schools don’t let students get their grades until they fill them out.
@Jakobian But this is a much stronger hypothesis than existence of second partials.
@Thor This is impenetrable. What about the Fermat cubic? Then we know there are 27 lines.
I agree it is impenetrable
This is classic stuff Fano knew how to do!
@TedShifrin Yeah, I know. I've worked at such institutions. I have... mixed... feelings about this.
22:47
I'm no Fano
@Jakobian Interestingly, Lang assumes continuity of $D^2f$.
23:00
oh, I just had an ephiphany
$\{L\vert L\cap X\neq\emptyset\}$ is always irreducible
so I just need to show that $\{L\vert L\subseteq X\}$ is closed and a proper subset
(the latter is obvious)
23:17
@Thorgott I'm silly, the latter condition, even in this case, is not polynomial in the Plücker coordinates
so this approach just doesn't seem to work
Mother Plücker!
23:59
Is it good time to say Merry Christmas yet?
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