Hi @TedShifrin, had some small questions, I'll add comments to what I've done for each (hopefully this isn't too long)
1) For question 12) b) from Sec 2.1 - you ask to show that the curve $\mathbf{g}(s_{0},t)$ corresponding to the function $\mathbf{g}(s,t) = \begin{bmatrix} \frac{st+1}{st-1} \\ \frac{s-t}{st-1} \\ \frac{s + t}{st-1} \\ \end{bmatrix}$
for this one I'm a little stuck. I don't know if there is another form of the expression to use, but based on what I know to show a set of points is a line I have to show that $\mathbf{x} = t \mathbf{u}$.
and the question which manifolds admit orthogonal coordinates
ok, im trying to figure out why homotopic maps induce isomorphic pullback bundles and am not quite seeing it. i was told this is easy. am i being stupid, was i lied to or both?
More precisely, I'm trying to show that the groupoid $\mathscr{B}G(X)$ of principal $G$-bundles over $X$ and isomorphisms is equivalent to $\Pi_1(BG^X)$. It seems like the right direction to try to construct my functor is $\Pi_1(BG^X) \rightarrow \mathscr{B}G(X)$, since given an actual $G$-bundl...
I think if $X$ is compact, I can always partition $[0,1]$ into subintervals $I_1,...,I_n$ such that each $X\times I_i$ is trivial and then proceed inductively
@dc3rd Well, the line doesn't go through the origin, of course. You can do it by algebraic manipulation. Perhaps less tricky is to show that $\partial\mathbf g/\partial t$ doesn't change direction. FYI, I like this question because it gives a parametrization of the hyperboloid where both the $s$-curves and the $t$-curves are the lines, but it's sort of a curiosity and not mainstream. Not at all essential.
@10Replies What does it mean to say that the $\mathbf v_i$ span $\Bbb R^n$?
The question @Thor alluded to (which I still do not really know the answer to) is: When does a Riemannian manifold admit charts so that $g_{ij}$ is a diagonal matrix?
Of course, I know the answer if it's a scalar multiple of the identity.
Obviously, no problem in dimension 2 and Deane Yang and Dennis deTurck proved in a paper that it holds in dimension 3 for every Riemannian manifold. I'm still uncertain what's going on in general.
@TedShifrin. Looking at the diagram from the text I had a feeling of that and felt that the lines would all have to be added to some "constant" vector because of just that.
If you're taking a linear algebra course, the first thing you need to do is write down a neat sheet of definitions. You must learn what the important terms mean. And you must know those, backwards and forwards.
You don't get to tell me you don't know.
@dc3rd: The "constant" vector depends on $s_0$, of course. You have to do some tricky division of polynomials if you're going to do it algebraically, but it's certainly do-able. But seeing that the direction vector doesn't change direction is easier.
@TedShifrin...yea I was looking at some sort of tricky division to try and isolate $t$ and it was looking ugly
Well the other two questions came from your P-Set 4:
A) c) verifying that the curve $x^{3} + y^{3} = 3xy$ which I parametrized in an earlier portion is symmetric.
B) Given the surface $z = f(x,y) = xy$ showing that the lines with direction vectors $\mathbf{u} = (1,0,b)$ and $\mathbf{v} = (0,1,a)$ and point $P = (a,b,ab)$ are enitrely contained in the surface.
for A) What definition are we using to define symmetry?
I had to go look at the problem. I said "the curve has an obvious symmetry." So I intended that you determine what it is, first of all. Doesn't the equation suggest something?
Problem B is related to the hyperboloid question. The saddle surface and the hyperboloid of one sheet are "the only" doubly-ruled surfaces other than planes.
@TedShifrin, I did....I was just thinking. Trying to figure out "what the equation suggests".........and also what it means to determine the symmetry. .........
The only thing that pops out to me about the equation is that it is close to but not exactly $(x - y)^{3}$....but I don't think this is your intention
I still think there's not enough time. :( But I guess partly the point is to get people to get a feel for everything so they can truly master it as they keep using it.
The definition of submersion is:
Let $f:U\to\mathbb{R}^n$ a differentiable function defined in the open $U\subset\mathbb{R}^m$, if for all $x\in U$, $f'(x)$ is an surjective linear transformation then $f$ is a submersion.
I have to find a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^...
Yes, of course. I think it's good advice regardless :)
@dc3rd: You're thinking too hard. Look how $x$ and $y$ appear in the equation.
@MikeM Of course they had to use them, too. :) I was always pleased a few people got all the points snd upset that some got only half (even with partial credit).
Hmmm....well then I need to go back and see the obvious symmetry then. So generally speaking then. The curve is symmetric through the line $y = x$. Is that what you mean by seeing the symmetry?
Well I'm not done yet.......I still have to verify this. So in terms of my parametrization I now would have $(y(t), x(t))$....but what needs to be verified?
So originally we were given $y = tx$.........reflecting my curve in the line $y = x$ changes my points from $(x(t),y(t)) \to (y(t), x(t))$. So we are doing this because we want to show my curve is symmetric along the line $y = x$. Being symmetric along $y = x$ means..........(gap)........which affects $t$ by.....(another gap)...........
ok...... @TedShifrin....I'm drowning here....what is it that I'm missing to be able to think this out more smoothly? Because this is sad right now and I'm beyond frustrated in myself
@Ted: Let $M$ be any element of $\mathfrak{so}(n)$. Then $T(Me_i, e_j, e_k, e_l) + T(e_i, Me_j, e_k, e_l) + T(e_i, e_j, Me_k, e_l) + T(e_i, e_j, e_k, Me_l) = 0$. Plugging $M = \delta_{ik} - \delta_{ki}$ gives $T(e_i, e_j, e_k, e_l) + T(e_i, e_j, -e_i, e_l) = 0$. So fixing $j, l$, $T(e_k, e_j, e_k, e_l)$ are all equal for $1 \leq k \leq n$. But summing over $k$ gives me trace, which is also zero, so these terms are all zero.
So in particular you get for example $T_{kjkj} = 0$.
It really does.......THis is going to suck, but I think I need to take a few weeks and go through all the fundamentals of geometry.........sigh.............
so you told me to draw a right triangle with slope $t$. The slope of this triangle is $t = \frac{y}{x}$. So to reflect this triangle would mean to switch my $(x,y) \to ((y,x)$. The slope of this reflection is then $t = \frac{x}{y}$.
Ok @TedShifrin.....this is upsetting, as you've observed my geometry is absolute garbage.....Any text in particular you suggest I should start with? Probably means going down to a simple high school text, cause this is sad........
I'll start with that. Thanks.....Going to be weird asking linear algebra questions about diagonalization, but here messing up basic geometry.....I'll be back in a few.
@zacts, finish that first. It will give you the framework of how to think about math questions. I'd suggest "How to read and do proofs" by Daniel Solow as well. Then make a run at Spivak
@TedShifrin: One second. Riemann curvature tensor of $S^2 \times \Bbb H^2$ is not traceless, right? I agree it has zero scalar curvature. But not Ricci-flat, is it?
By traceless tensor I meant $\sum_i T_{ijik} = 0$, for all $j, k$. This is what Weyl tensor satisfies.
I'm struggling to see why this is so complicated, can I not just say, because v_1 .. v_p spans R^n and T(v_i) = 0 for all I then T(X) = 0 because x is in R^n
@10Replies I don't know the Row-Pivot Theorem, but if $\{v_i\}_{i=1}^p$ span $\mathbb{R}^n$, then for any $x\in\mathbb{R}^n$, there are $\{c_i\}_{i=1}^p$ so that $x=\sum\limits_{i=1}^pc_iv_i$.
is the product of conformally flat things always conformally flat by this theorem or are there counter-examples that just need to be more complicated (I imagine non-constant curvature)?
Actually maybe I still don't know if $S^2 \times \Bbb H^2$ is a counterexample or not.
Certainly the metric is diagonal, but does it admit "enough diagonal coordinates" in the sense of DeTurck-Yang?
$W_{ijkl} = 0$ only if there are enough diagonal coordinates. I think I missed this subtlety when I worked it out before
Just try to compute the Weyl tensor explicitly.
It's probably nonzero.
Ted has to be right, I mean, it looks like a totally overdetermined system. The point is I think that you cannot expect diagonal coordinates with frame at a point not respecting the product structure at all.
@Thorgott @Ted @Ryan Simply $S^2 \times S^2$ is not conformally flat I think. It has varying sectional curvature (take (geodesic) x (geodesic), these are isometrically embedded flat torii). But it is also clearly an Einstein manifold, those are conformally flat iff they have constant sectional curvature.
Follows from writing down the Weyl tensor.
So yeah, differential geometers are correct and I am wrong. Sorry!
Take a point $p$; then there's a geodesic flat torus passing through it, so the tangent plane $P \subset T_p M$ to that torus has sectional curvature 0 right?
But take the $S^2 \times \{*\}$ factor, which has sectional curvature 1
Guys, would anyone mind having a look at this question? https://math.stackexchange.com/questions/4010742/sheaf-of-1-forms-on-affine-opens I had to type out a few results and a commutative diagram, hence felt it was better to write it out on the main
Question boils down to understanding a result (without proof) on restriction maps for a sheaf of 1-forms of an algebraic variety between affine opens
@Balarka why are characteristic classes fucking witchcraft, man
I thought "let's put an arbitrary connection on this bundle and take the characteristic polynomial of the curvature form" or whatever is incomprehensible, but the algebraists definition is incomprehensible too
@Thorgott They're not really. The differential geometric approach is inspired by the case of complex line bundles, where the fact that the result is independent of choice of connection is a result of Hodge theory. But I tend to think that inspires one to study invariant polynomials in curvature and not characteristic classes.
The definition of characteristic classes are just cohomology classes natural under pullback and preserved under direct sum. That's it! They're gadgets to tell whether or not a vector bundle is stably trivial.
From which one concludes, ok, I just need to understand the universal case of cohomology classes on BO. You study the cohomology of that space by whatever techniques and find out it's a polynomial ring, and then you realize you may as well try to understand the generators.
From which you get SW classes which you understand and work with axiomatically.
For some reason people think it's a good idea to try to understand these straight from the definition. But the definition is just a recipe to pull out these generators.
If I were really trying to give a definition I would do it geometrically. w_1(E), E rank n, is the generic locus where n sections become linearly dependent. (Aka where a section of the determinant bundle generically vanishes.)
That literally measures triviality of determinant line bundle, hence obstructs orientability.
w_n(E) = mod 2 Euler class is where a generic section is zero.
And w_j(E) is where (n-j+1) sections generically become linearly dependent.
This is a geometric interpretation of the obstruction theory definition: they are the first things you see stopping you from finding linearly indep sections.
The last four or five messages are almost certainly literally wrong but morally right
I don't know the classifying spaces viewpoint (perhaps I should), but I think that should amount to saying that each $n$-bundle admits a unique homotopy class of bundle maps into the tautological bundle over the infinite $n$-Grassmannian and that Stiefel-Whitney classes come from pulling back the cohomology of the infinite Grassmannian along this. That approach makes sense to me conceptually. But how do I fit Chern classes into this framework?
Is the power set of $\{a,\{a\},\{\{a\}\}\}$ the set $\{\emptyset, \{a\}, \{\{a\}\}, \{\{\{a\}\}\}, \{a,\{a\}\}, \{a, \{\{a\}\}\}, \{\{a\},\{\{a\}\}\},\{a,\{a\},\{\{a\}\}\}\}$ ?
Also, how do we make the "generic locus" idea precise? I've only heard of stuff like generic points in the context of alggeo (and not that I know them there either).
@MaryStar Simply because it isn't true. If you think it is true, you can tell me why you think so and I can tell you why you are mistaken. In this case, it depends on what your $a$ is.
If $a=\emptyset$, it's true, I think it's false for any other $a$, but I haven't checked the details
@Mike Yeah, I wouldn't have known a priori. The definition of Chern classes I have is as elegant as it is intransparent. If they just come from the complex Grassmannian, that's very clear conceptually (though I guess that means I should try understanding why that's equivalent to my definition).
also tranverse section = transverse to the zero section?
Let $E\rightarrow Y$ be a complex $n$-bundle and form the external tensor product bundle with the tautological line bundle over $\mathbb{CP}^n$ to obtain a bundle over $X\times\mathbb{CP}^n$, whose Euler class is an element of $H^{2n}(X\times\mathbb{CP}^n)$. Via Künneth this is isomorphic to $\bigoplus_{j=1}^nH^{2j}(X)$ (where we identify the cohomologies of $\mathbb{CP}^n$ with $\mathbb{Z}$ canonically via the generator that is the appropriate power of the Euler class of the tautological line bundle) and the component of that Euler class in $H^{2j}(X)$ is the $j$-th Chern class.
The way one might say this is that if you take a bundle E and projectivize to P(E), Leray Hirsch gives you that its cohomology is a free module over H^*(M), isomorphic to H^*(M) o H^*(P(E_x)) for some fiber E_x. But this isomorphism doesn't preserve the ring structure. The c_i show up in the formula for the ring structure.
The definition of submersion is:
Let $f:U\to\mathbb{R}^n$ a differentiable function defined in the open $U\subset\mathbb{R}^m$, if for all $x\in U$, $f'(x)$ is an surjective linear transformation then $f$ is a submersion.
I have to find a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^...
