But yeah if I remember right, the way I had said my solution was that the set of polynomials with prime power order at most $n$ was a vector space of dimension $r$, but that the ideal generated by a given polynomial $f$ capped at degree $n$ was another vector space of dimension $k$
@TheGreatDuck In any case, odd-degree polynomials are minus infinity at the left and plus infinity at the right, so they have to hit zero at some point
I didn't mean to act like I cared to know where you live Duck. I just think that you were the one pretending to be Balarka. I don't want you to think I'm some sort of weirdo or anything. I honestly just think I am a detective.
@Dodsy fair enough. Is there any way we could talk in a private room by ourselves on here? There's something I want to tell you that might be relevant.
Also, guys, say we want to factor $u^2+v^2$. We could do this in two ways: $$ (u+v)(u-v)\quad\text{or}\quad(iu+v)(-iu+v). $$ When would one prefer the second factorisation? I'm guessing when $u$ and $v$ are complex, but I'm not sure.
@Danu: Yes, the generic Riemannian manifold has no isometries. Obviously, you're working with homogeneous spaces, but you have to be very careful about how homogeneous they actually are.
I've given up on Queens Ted. The manager just sent the email back to the guy I've already been dealing with, who will probably be angry that I tried to go over his head now. Thus, I am relying solely on UWO.
I haven't thought about it, @MikeM, but likely true. I know that symmetric spaces are much better than general homogeneous spaces for purposes of invariant cohomology (on a general homogeneous space, invariant forms aren't even necessarily closed). I haven't pondered this aspect.
@TedShifrin So, I realized something quite intriguing in response to that surface geometry thing we were discussing the other day. Were you aware that partially folding a plane preserves the rules regarding opposite and adjacent angles between intersecting lines?
@TedShifrin decided to try moving objects along the polygonal 3D model instead of a smooth curve. Locally everything is Euclidean so it's mostly trivial.
What's the name for a manifold that looks the same in every direction? (At every point there's an isometry taking one tangent vector to any other at the same point?)
I just realized that Salamon Appendix D is a 5-page proof that (if int f, int h are positive, and $h$ is weakly positive in that you can integrate a nonnegative test function against it and always get a nonnegative answer) you can solve the equation $\Delta u + e^u h = f$, which should be enough to get Kazdan-Warner's results for surfaces
@Danu I haven't been posting mathjax. there's a special symbol "" that can be posted and literally renders to nothing in the browser. I can post blank messages and it's kind of funny to see what it does. I've been posting a few tests every now and then to see how it interacts with mathjax.
Salamon solves that eq'n for all compact n-manifolds, which surely works into the general proof of their theorems, but the change of conformal class formula for scalar curvature involves a |df| in higher dimensions
