If $\{U_i\to X\}$ is an etale cover, and $\mathcal{F}$ is a sheaf for the etale site $X_{et}$, how would you interpret $$\mathcal{F}\otimes_{\mathcal{O}_X}\mathcal{O}_{U_i}?$$ Would you take it to mean $\mathcal{F}\otimes_{\mathcal{O}_X}f_*\mathcal{O}_{U_i}$ where $f:U_i\to X$ and the $\mathcal{O}_X$-module structure given to $\mathcal{O}_{U_i}$ given by $f^\#:\mathcal{O}_X\to f_*\mathcal{O}_{U_i}$?
The context was a seminar talk (which I have bad notes for), but I had found similar uses in an etale cohomology text, let me see if I can find a specific one
That being said, I can't see any other way to interpret it atm
Two alternative questions for you. Are you familiar with the formalism of ringed sites? Does the structure sheaf $\mathcal{O}_X$ or the structure sheaves $\mathcal{O}_{U_i}$ give etale sheaves on the site $X_{et}$ in a natural way? Rather than these just being zariski sheaves on their appropriate zariski sites
I'm looking at page 190 of Lei Fu's etale cohomology, looks like we turn a quasi-coherent sheaf $\mathcal{M}$ of $\mathcal{O}_X$-modules into an etale sheaf on the site $X_{et}$ by taking $U\mapsto \Gamma(U,\pi^*\mathcal{M})$ for any $U\to X$ in $X_{et}$, where in particular we take $\mathcal{O}_X=\mathcal{M}$ to get $\mathcal{O}_{X_{et}}$
Given an $S$-scheme, $f:Y\to S$, such that the structure morphism $f$ is affine, and a sheaf $\mathcal{G}$ on $S$, by making use of the isomorphism $H^i(Y,f^*\mathcal{G})\cong H^i(S,f_*f^*\mathcal{G})$, we can get a well defined homomorphism $H^i(S,\mathcal{G})\to H^i(Y,f^*\mathcal{G})$ by making use of the unit of adjunction applied to $\mathcal{G}$, i.e. we have a morphism of sheaves $$\mathcal{G}\to f_*f^*\mathcal{G},$$ which induces a morphism: $$H^i(S,\mathcal{G})\to H^i(S,f_*f^*\mathcal{G})\cong H^i(Y,f^*\mathcal{G})$$
That you also answered, but your remark seemed slightly cryptic at the time
I guess I'm just having one of those days where I feel retarded, and nothing makes sense
Is it valid to define the set of dual vectors $f_i$$^j:(1,1)\rightarrow\mathcal{C}$ such that $f_i$$^j(T) = T_i$$^j$ for any $T\in\mathcal{T}_1$$^1(V)$?
Seems like I need that set of duals in order to define the "trace" of a $(1,1)$: $\text{Tr} \equiv \sum_if_i$$^i$ $\Rightarrow$ $\text{Tr}(T) = \sum_iT_i$$^i$
I mean, it doesn't make intuitive sense that by increasing the upper sum limit by multiplying by n, you also multiply the n terms in the formula it is equal to.
How did you tell at a glance that the form from yesterday was closed? I computed its exterior derivative and got zero, but I doubt that's how you decided it's closed
Yes, that's how you decide. I knew it because I know it. :) If you think it through in spherical coordinates, it's clear. In 3D, you have $dA$ for the $2$-sphere is $\rho^2\sin\theta\,d\theta\wedge d\phi$ (using European coordinates), and you're dividing through by $\rho^2$, so what's left has no $\rho$ dependence (we already discussed that), and so taking $d$ you have to get $0$.
So I have $S^{n-1}$ as a Riemannian manifold with the metric and orientation it inherits from $\Bbb R^n$ with the standard ones. I want to verify that the volume form on $S^{n-1}$ is $\omega=|x|^{-n}\sum_{i=1}^n x^i dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n$
You can also make a local argument with Stokes's Theorem, based on my observation that the flux across (i.e., integral over) spheres centered at the origin is independent of radius.
@ConstantineBlack If you've ever graphed data on log-log paper, you're computing the slope in that plot.
I'm saying you can make a Stokes's Theorem argument on a little region. If $d\omega\ne 0$ at some point, then it won't change sign nearby. But you can choose the region to be a "spherical" frustum (flat sides on which $\omega=0$ with little pieces of spheres for ends).
@TedShifrin Thank you. I figured out, there was a typo in the notes I was checking and I couldn't make sense of the result I was getting using the formula and what the notes where having. Thanks again.
Now I'm asked to prove that the divergence of a vector field on an orientable Riemannian manifold is independent of the orientation, but it's not clear to me what exactly that means. I can write down a local coordinates expression for $\mathrm{div} X$ depending only on $X$ and $g$ but not on the volume form, isn't that enough?
@Alessandro I haven't heard anything back from that prof but my friend is doing his bachelorarbeit under him and said that he's really bad at replying rofl
@TedShifrin Ah, I guess not because the components of $X$ might change sign in local coordinates if my new charts have the standard basis for $T_p M$ oriented differently
@AlessandroCodenotti yes for me is also right!! Now my problem is to translate "Conjuncts" to Spanish (my native language)... :D but don't worry, you've done your work
Now if we have $x\vee y$, $x$ and $y$ are disyuntors?
@AkivaWeinberger yes, but what is the translation for "Conjuncts"? Conjuncts = Conjunciones, so then a Conjucnt will be a Conjunción, which is equal to Conjunction
When students explained to me that they had a professor across campus who traditionally kept them late, I told them not to worry, just to do their best. But oversleeping is NOT a valid excuse.
Although I never required attendance, I usually was aware when students were missing (since I had 30 or fewer students, typically) — and since I like a lot of class participation, having students missing might have a deleterious effect.
Sometimes books just don't do an adequate or tasteful job. Of course, when I wrote my own I could no longer complain, but I still did a lot of examples and heuristics that weren't in the book.
I try to put myself in the professor's shoes. If they didn't show up, there might be students who would want to learn something but can't. Class topics would fall behind, it would just be bad all around. And if the professor has to come to class, why should I skip out?
I guess if you're going to be rude and not pay attention/participate, I'd just as soon you were not there. Having said that, if you've got a high grade in my class and I know you're bored, I will tend to let it go. ... Of course, this is all moot now, since I've retired.
I have a representation of a Lie algebra that looks like $(\Bbb Z/2\Bbb Z)^N$ in the sense that all states are of the for $r(q_{i_1}q_{i_2}\dots q_{i_k})x_0$ for a ground state $x_0$ with $q_{i}^2 = 0$ and $[q_i, q_j] = 0$ and there are $N$ such $q_i$. How do I word the similarity with $(\Bbb Z/2\Bbb Z)^N$ ?
In one direction it seems reasonable. I can use the $R\otimes_{\Bbb Z}S^{op}$-module structure to define a left $R$-module structure, and a right $S$-module structure, such that they are compatible
To be more precise, given a left $R\otimes_{\Bbb Z}S^{op}$-module $M$, I can define left $R$-module structure on $M$ by taking $r\cdot m = (r\otimes 1)\cdot m$, and left $S$-module structure by $m\cdot s = (1\otimes s^{op})\cdot m$
Of course I'd have to write it all up to be rigorous. For the other direction I want to take an $R-S$-bimodule $M$, and give a left $R\otimes_{\Bbb Z}S$-module. So a potential choice is to take $(r\otimes s^{op})\cdot m := r\cdot m \cdot s$
Okay, your first assertion is that a left $R$-module $M$ is given by $R\to \text{End}_{\Bbb Z}(M)$. Where I suppose you send $r$ to the endomorphism it defines $r\cdot -$, which is $\Bbb Z$-linear since all abelian groups are $\Bbb Z$-modules via $n\cdot x = x+\cdot +x$, and $$r\cdot (x+\cdots+ x)=r\cdot x +\dots r\cdot x= n\cdot (r\cdot x)$$ and similarly for a right $R$-module
so we need a tetralinear(?) map $A \times B \times A \times B \to A \otimes_\Bbb Z B$ given by $(a_1, b_1, a_2, b_2) \mapsto (a_1 a_2 \otimes b_1 b_2)$
if you want to use universal properties, I'm taking the product of the two maps to get $A \times B \to C \times C$ and then composing it with the ring multiplication $C \times C \to C$
You were taking $M$ to be a left module with respect to both $A$ and $B$ (i.e. $C=\text{End}(M)$ and $A,B\to C$ are both giving left module structure)?
@jcora after clicking into your profile to see the picture enlarged, it's clear you aren't, but in the little picture, it looks like you're holding an ak-47
we usually have examples in R^n -> R^m, wasn't sure how to approach this one. I'll try it out to see if I can understand treating R^n x R^n as R^2n in this case
@TedShifrin don't you need a candidate solution to place it in the limit, and see if =0? and then conclude that this is the unique linear operator? I didn't see an explicit definition of Df that would let me compute it directly
neat! we usually defined directional derivative as $\partial_v f(x, y) = \nabla f(x, y) \cdot v$, and this definition makes sense to me too, but I've never seen it before
