Yeah. Oh well. I guess I just like to apply that fact ($H^n(X; G) \cong [X, K(G, n)]$) everywhere, and it looks plausible since skeletons of $K(\Bbb Z, 2)$ are actually manifolds.
I am still not sure why I can choose $g$ to be smooth.
No problem! I happened to have studied this stuff last year (and partly this year) so I'm glad I can put it to use. If you have any other questions while I'm on, it's good practice. ^_^
@Ted: Not sure that's more interesting. You're fixing a Hermitian structure on a complex manifold and want it to be locally, but not globally, Kahler? Metrics and complex structures being compatible is a local condition.
TIL: Calabi-Yau manifolds are algebraic. Adorable proof.
@Balarka I'll look at it more later, but my initial thought in the first paragraphs: couldn't we scale the three pieces of the tribar independently without changing our perception of it? That would mean a symmetry group of ${\Bbb R}_{>0}^3$. What's instrinsically special about the subgroup of things that fix the third piece and whose three scaling factors multiply to one?
@anon No, I do not think any more scalings are possible. The deal is that if you enlarge the tribar by $\lambda$ and take it $\lambda$ far away, then the tribar looks exactly the same.
Careful analysis of the Weitzenbock formula shows that harmonic $(p,0)$-forms are the parallel $(p,0)$-forms. (Here you use Ricci flatness. You need (p,0) for the Weitzenbock formula to be of much use.) Then we've reduced calculating $H^{p,0}$ to a linear algebra problem: what are the fixed subspaces of $\Lambda^p T^{(1,0)}^*$ under the action of $SU(n)$?
Yeah, I'm finding calculus very interesting so far. I think I like limits the best, because it's less straightforward that derivatives (at my current level) but not as tricky as integrals.
This is not hard to calculate; we get $b^{p,0}=1$ for $p=0,n$ and 0 otherwise. In particular, if the dimension $n>2$, $b^{2,0}=b^{0,2}=0$. So $H^2=H^{1,1}$.
@TedShifrin Yeah, I've heard quite a bit about that! Maybe that can keep me busy since I'll have a fairly large head start when I actually take a calculus class next year.
But we're Kahler!! In fact the Kahler cone is open and nonempty. So I can pick a rational Kahler form. This is automatically a positive (1,1) form - taking a suitable power of it gives me a line bundle represented by a positive (1,1)-form.
@anon "scaling the pieces at all changes the tribar" not true. think about sclaing one piece by $1$, other by $\lambda$, and another by $1/\lambda$. The figure would be the same. Note that you're looking at the projection, the tribar is not really sitting in R^3.
I don't know what "perception" means, to be honest.
The fact here that the dim is at least 3 is essential because there are non-algebraic K3s. I find this whole thing (including the automatic "explanation" of why it doesn't work for K3) so charming.
@anon the tribar (the 2d picture) is projection of three pieces $Q_1, Q_2, Q_3$ sitting in $\Bbb R^3$, appropriately aligned. Enlarge $Q_1$ by $\lambda$ in $\Bbb R^3$, $Q_2$ by $1/\lambda$ in $\Bbb R^2$ and leave $Q_3$ alone. Project again. The projected tribar would not change.
if you scale $Q_1$ by $\lambda_1$, and $Q_2$ by $\lambda_2$, and $Q_3$ by $\lambda_3$, then the resulting pieces each look like $Q_1,Q_2,Q_3$ respectively still no?
Then the usual theorem is that a Kahler manifold with c_1=0 has a metric making that true. I don't know if it's true that a complex manifold with c_1=0 must be Kahler? But I bet no?
@TedShifrin Yeah, I intend to do that. I like to know why things work, rather than just plugging stuff in. Something I really dislike about my current math class is that there's absolutely no emphasis on why we do things or what it means
That's a shame, @TedShifrin, I've heard good things about the calc teachers though so I'm hopeful. According to one of my teachers last year, one of the calc teachers does a lot of math in his free time so maybe he'll teach (or be available to explain) more of the details/concepts
@Walker23: generally, it's all about training the students to get 5's on the exam. But some teachers push way past that. (I taught 36 years of students who'd taken AP calculus of varying qualities.)
@MikeM, I've forgotten all the appropriate definitions, sadly, even though I once knew them very well. But I need to go finish packing and try to get some sleep (at this insane hour).
Appropriately I learned them in the past week. Having fun with this, though it's time for me to put down the holonomy for now. Have a good and safe flight!
@anon Say you enlarge $Q_1, Q_2$ by $\lambda_1, \lambda_2$. $Q_1', Q_2'$ be the resulting pieces. Then it need not be the case that $Q_1'$ and $Q_2'$ "match up" from the place we are looking at, whereas $Q_1, Q_2$ did. So projecting the resulting thing down need not even give a tribar.
Pick the piece $Q_2$. Assume it's $\lambda$ times far away from where you thought it is. Then $d_{12}$ gets multiplied by $\lambda$, $d_{23}$ by $1/\lambda$, and nothing happens to $d_{31}$. So automatically, changing the "perception" (I hope I understand that word of yours correctly) of $Q_2$ by $\lambda$ changes perception of $Q_1$ by $1/\lambda$ and leaves the perception of $Q_3$ unchanged.
So no other symmetries are possible.
I think this is the same as what I meant by stretching. You're simply moving away from where you were instead of physically stretching each piece.
I haven't really thought about formalizing the ambiguity group, and I don't think there's a straightforward way to do that. Most of my answer consist of formalizing the cohomology group.
@anon Here's an even better explanation. Take $Q_1, Q_2, Q_3$ to be the three bars here, appearing in the same order. Take $Q_1$ $\lambda$ far away, $Q_2$ the place it is, and $Q_3$ $\lambda$ closer. This is the same as stretching $Q_1$ by $1/\lambda$, $Q_2$ by $1$ and $Q_3$ by $\lambda$. The resulting tribar is the same however. If you used arbitrary $\lambda_i$'s as scaling factors, you could never do that.
@DanielFischer: on wiki, the definition for a Radon measure is that it is inner regular and locally finite. in the standard solution to one of my exercises, they state "... is a Radon measure, i.e. inner and outer regular". is this equivalent? I don't see why it should be, in general.
@Huy On e.g. $\mathbb{R}$, the measure $$\mu(A) = \begin{cases} 0 &, A = \varnothing \\ \infty &, A \neq \varnothing \end{cases}$$ is inner and outer regular (at least, with the definitions of regularity I recall). It's not locally finite, however. So the two aren't equivalent. They may be equivalent in the presence of other, unstated, conditions. Anyway, it does happen that different people use the same word for different things. That's normal.
Another definition of a Radon measure, iirc, is that a Radon measure is a continuous linear form $\mu \colon C_c(X,\mathbb{R}) \to \mathbb{R}$.
@iwriteonbananas I just got to know yesterday that every continuous map between smooth manifolds is htpic to a smooth map. Thus, the htpy $M \times I \to N$ is homotopic to a smooth homotopy. But can I choose this homotopy to be relative on a subset? That is, if $f : M \to N$ is continuous, $f$ smooth on $W \subset M$, is $f$ homotopic to a smooth map $g : M \to N$ relative to $W$?
I'd want to see an answer to this. Most of the time people works with projective resolutions instead of free ones, so I'm curious why wiki does it with free things.
The parametrization of a torus is $\sigma (\theta, \phi)=((a+b\cos\theta)\cos\phi, (a+b\cos\theta)\sin\phi, b\sin\theta)$, right? Which is the interval of $\theta$ ?
Hello, is there an example of a field k and automorphism f on k[X_1,...,X_n] so that f is identity on k but does not preserve the total degree of a polynomial?
I have found that the Gauss curvature of the torus is $K=\frac{\cos\theta}{b(a+b\cos\theta)}$. Since $b<a$, we have that $K>0 \Leftrightarrow \cos\theta >0 \Leftrightarrow 0<\theta<\frac{\pi}{2} \land \frac{3\pi}{2}<\theta<2\pi$ and $K<0 \Leftrightarrow \cos\theta <0 \Leftrightarrow \frac{\pi}{2}<\theta<\frac{3\pi}{2}$.
Thanks, @BalarkaSen! Yep, I recalled it from geography lessons which are now a distant memory, @Huy :-b
That seems fine, @MaryStar. Might be more explicit (possibly less rigorous) to say that $$ b\cos \theta \in [-b,b] \overset{a>b}{\implies} a + b\cos\theta \in \left[a-b,a+b] $$ where $a-b > 0$.
"Another nice feature of the SVD is that the number of nonzero entries in the diagonal matrix is equal to the rank of the matrix. In this way the SVD is said to be rank revealing. Thus $r(P_X) = r(X)$." Can someone point me to where I can find a proof of this? ($X$ is a $n \times p$ matrix with real entries, $P_X = X(X^{T}X)^{-}X^{T}$ is a projection matrix of $X$ - notice here that I'm using the pseudoinverse, rather than $(X^{T}X)^{-1}$)
@Huy Other than $X = U\Sigma V^{T}$, no. I'm not sure how to extract the rank from this given the information. [Sigh, I've forgotten how to get rank if I don't have the elements of the matrix in front of me]
it is defined as the dimension of the image, usually
maybe you also defined it in terms of row rank and column rank and later showed the two are equal and also equal to the dimension of the image of the linear map induced
I would like to prove $r(X) = r(P_{X}) = \text{tr}(P_X)$, $r$ denoting the rank, $X \in M_{n \times p}(\mathbb{R})$, and $$P_{X} = X(X^{T}X)^{-}X^{T}$$
where $(X^{T}X)^{-}$ is a generalized inverse of $X^{T}X$.
I don't believe I know the machinery to prove this. But here's what I do know that mi...
@Huy I can compute the trace of $P_X$ easily, given the information from the answer. Why is this equal to $r(P_X)$, and how do I know that $r(X)$ must be equal to $r(P_X)$?
Sure, I get that. The trace is going to be $r$ (using the notation from the answer). But why is this going to be the rank of $P_X$? I'm missing something here. @Huy
The comment was "Another nice feature of the SVD is that the number of nonzero entries in the diagonal matrix is equal to the rank of the matrix." Why is this?
@Huy Well, that's the funny part. In stats, we say that a least squares estimator for the estimates is $\mathbb{E}[Y]= X(X^{T}X)^{-}X^{T}\mathbf{y}$ and then we for some reason call $X(X^{T}X)^{-}X^{T} = P_{X}$. I know nothing about $P_X$ beyond this and the properties of $P_X$
@Clarinetist: I need to leave soon but I can probably help more tonight or tomorrow if you haven't resolved it. I'm not 100% sure yet if $P_X X = X$ solves all of it, but I think it does.
maybe you can figure it out yourself in the meanwhile
basically all that's left is to see that $P_X$ and $X$ have the same image. from $P_X X = X$ you already know that the image of $P_X$ contains the image of $X$ (if I'm not thinking too quickly).
