Let's just see what happens to $f \otimes x_2 \wedge \cdots \wedge x_k \in S W \otimes \wedge^{k-1} W$ under this weirdness. It first goes to $\sum_{i = 2}^k x_i f \otimes x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge x_k$, in $S^2 W \otimes \wedge^{k-2} W$. That goes to...
$\sum_{i = 2}^k \sum_j \partial_j (x_i f) \otimes x_j \wedge x_2 \wedge \cdots \wedge \widehat{x_i} \wedge \cdots \wedge x_k$, again in $SW \otimes \wedge^{k-1} W$. How do I simplify this?
ted, i've noticed the same thing. (question from something fairly advanced with lots of very un-advanced confusion.) it's baffling. i could sometimes see it if it's chapter 1 material in a grad course, because who knows where people came from before that. but it's usually more advanced.
i blame the internet for making it way too easy to do the equivalent of, hm, what's that? seems interesting. i think i'll get 20 graduate textbooks on it.
@TedShifrin It was much nicer to have her home for her passing. She was as comfortable as we could make her and I was cradling her head and petting her neck and ear and her mom was petting her snout. She was licking her mom's hand, which always calmed her.
duck pond was something of a bust today. she complained about having to walk around, then saw that someone brought their pet rabbit to the park. that was all she wanted to look at. then she asked where the dead heron from last week was.
she is almost normal on flat surfaces but her foot really juts out at a weird angle when she's walking on grass. i don't understand how it could make too big a difference, but it does.
Here is what I think is correct. Let's call $d_K : W \otimes \wedge^{k-1} W \to S^2 W \otimes \wedge^{k-2} W$ the Koszul differential and $d_E : S^2 W \otimes \wedge^{k-2} W \to W \otimes \wedge^{k-1} W$ is the exterior-like differential. Then $d_E d_K - d_K d_E$ is identity, or some multiple of it
Last I did the computation I got some multiple which messed things up because characteristic
it seems like you're trying to do some kind of russell's paradox. the gamma of all gammas and supserscripts and subscripts that aren't superscripts or subscripts of themselves.
Here is the confusing calculation, if you care (I will do this with general $n$): We use all the identifications and notations from earlier. The map $f : \Bbb R^n \to \Bbb R$ gives derivative $Tf : T\Bbb R^n \to T\Bbb R$, $Tf(x, u) = (x, Df(x)u)$. Taking derivative again, and using $T^{(2)} \Bbb R^n = \Bbb R^n \times T_0 \Bbb R^n \times T_0 \Bbb R^n \times T_0T_0 \Bbb R^n$, I get $Tf(x, u_1, u_2, v) = (x, Df_x(u_1), Df_x(u_2), D^2f_x(v, u_2))$. Now, using
ted at the moment we're thinking a red dress or shapeless felt thing outfitted with some yellow 'seeds,' some sort of green hat, and (to make it scary) a cape from her black cat costume that's got spider webs on it.
Alright if everything checks out I'm getting that if $d_{i, j} : S^i W \otimes \wedge^j W \to S^{i-1} W \otimes \wedge^{j+1} W$ is the Koszul differential and $\partial_{i, j} : S^i W \otimes \wedge^j W \to S^{i+1} W \otimes \wedge^{j-1} W$ is the exterior-like differential, then $\partial_{i-1, j+1} \circ d_{i, j} + d_{i+1, j-1} \circ \partial_{i, j} = (i + j) \mathbf{1}$.
We'd like to prove $\ker(d_{i, j}) \subseteq \mathrm{im}(d_{i+1, j-1})$ using this, I suppose
I guess one can try to cook up a counterexample over $\Bbb F_p$. Maybe with $W = \Bbb F_p^2$. If it goes wrong it has to go wrong at $0 \to S^{p-2}\Bbb F_p^2 \otimes \wedge^2 \Bbb F_p^2 \to S^{p-1} \Bbb F_p^2 \otimes \wedge^1 \Bbb F_p \to S^p \Bbb F_p^2 \to 0$?
Because $p-2 + 2 = p-1 + 1 = p+ 0 = p$, not invertible
The ironic thing is that he corrected a bunch of things for this particular homework a week ago and sent a revised version, added this problem to make up for the wrong ones
So if you have a sequence of isometries $f_n$ with $f_n(p) \to q$, I want to prove this sequence has a subsequential limit isometry -- this is exactly what properness means.
Don't think so. Properness of action of $G$ on $M$ means $G \times M \to M \times M$, $(g, m) \to (gm, m)$ is proper, not that $G \times M \to M$ is proper, remember
@TedShifrin the motivation for asking about the hyperbolic case was that the Riemannian manifold was that the Riemannian manifolds for which I have the complex analysis argument happen to have a hyperbolic metric
@Ted: So just to nail the coffin since I'm writing the solution: For $M = \Bbb F_2^2$, the exact sequence $0 \to \wedge^2 M \to S^1 M \otimes \wedge^1 M \to S^2 M \to 0$. The first arrow maps the basis vector $x \wedge y$ to $xy \otimes 1$, that's that. The image of this does not contain $x \otimes y + y \otimes x$, which maps to by the second arrow $xy + yx = 2xy = 0$.
One second, first arrow wrong.
$x \wedge y$ maps to $x \otimes y + y \otimes x$, actually, no?
Bit confused
Oh, maybe I should look at $(x + y) \otimes (x + y) - (x \otimes x) - (y \otimes y)$ in the middle term. This isn't in the image of $\wedge^2 M \to S^1 M \otimes \wedge^1 M$, which is spanned by $x \otimes y + y \otimes x$
Still it maps to $(x + y)^2 - x^2 - y^2 = 2xy =0$?
@BalarkaSen what exactly is this exact sequence? How do you have a map $\wedge^2 M \to M \otimes M$ with no assumptions of char $0$ which allows for antisymmetrization?
Then I don't see how the sequence you wrote can fail to be exact, at least when $M$ is a finite-dimensional vector space. The fact that $M \otimes M \to \mathrm{Sym}^2 M$ is surjective and has kernel generated by $x \otimes y - y \otimes x$ is just the definition of $\mathrm{Sym}^2M$. So the only thing that could go wrong beyond that is that $\wedge^2 M \to M \otimes M$ is not injective. But that's impossible by dimension counting
@BalarkaSen if you work with symmetric tensors, not with $M \otimes M / \langle x\otimes y -y\otimes x\rangle$, then I don't see the map $M \otimes M \to S^2 M$
$M$ is a free module in my case. $S^2 M$ is degree 2 homogeneous polynomials
So you just send a tensor to the corresponding homogeneous polynomial
It seems right that $0 \to \wedge^2 M \to M \otimes M \to S^2 M \to 0$ should be exact, but it sounds wrong that you can symmetrize a tensor in characteristic 2
Don't really have the brain power to think this through today
@BalarkaSen I don't think that the exactness of $0 \to \wedge^2 M \to M \otimes M \to S^2 M \to 0$ directly gives you a way to symmetrize tensors. That would require you to have a splitting $S^2 M \to M \otimes M$
went out for drinks in berkeley tonight. ended up at eureka! near the berkeley bart station. one of our party forgot her id so we couldn't get into tipper & reed.
Here’s a group theory result which will lead to what you want. If $G$ is a finite abelian group, let $m$ be the largest order of its elements. Then the order of every element divides $m$.
