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Apr
16 |
comment |
A finite supersolvable group with generators of prescribed order
Usually, two-generators groups are well-suited for being represented on a surface. You may look for keywords like "hyperbolic groups" or "Fuchsian group". You use the Cayley graph and the natural action of the group on it (see, e.g., sciencedirect.com/science/article/pii/0095895683900096 ). In the most cases, the group is infinite (see the Burnside problem). |
Apr
15 |
accepted | How to embed $S^2\mathbb{C}^2$ into $S^2S^3\mathbb{C}^2$ and get the ideal of the twisted cubic? |
Apr
14 |
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How to embed $S^2\mathbb{C}^2$ into $S^2S^3\mathbb{C}^2$ and get the ideal of the twisted cubic?
Very concise and precise explanation - thanks! However, I had no doubts that the space which I called $\ker p$ identifies with a symmetric square. I wanted to see explicitly the identification, and understand to what extent is natural. As you can see from the comment above, I followed your reasoning, and I wrote down the identification. But I still can't see why it is an $\mathfrak{sl}_2(\mathbb{C})$-module morphism. |
Apr
14 |
revised |
How to embed $S^2\mathbb{C}^2$ into $S^2S^3\mathbb{C}^2$ and get the ideal of the twisted cubic?
added a comment after Sasha's answer |
Apr
14 |
asked | How to embed $S^2\mathbb{C}^2$ into $S^2S^3\mathbb{C}^2$ and get the ideal of the twisted cubic? |
Apr
14 |
accepted | Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian? |
Apr
13 |
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Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
Interesting point. Incidentally, are you claiming that $\mathbb{P}T^*S^3=\mathrm{O}(4)/\mathrm{O}(2)$? Or, stated differently, that there is an $S^1$-bundle over $\mathbb{P}T^*S^3$ isomorphic to $\mathrm{O}(4)$? |
Apr
13 |
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Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
Again, you're right: the Killing form does not induce a metric, but rather a conformal class of them. Nevertheless, I could begin with the Euclidean metric on $S^3$, i.e., the restriction of the Euclidean metric on $\mathbb{R}^4$. I keep suspecting that $\mathbb{P}T^*S^3$ is equipped with a distinguished family of contact sub-Riemannian structures, and that this family is of rank two (i.e., described by the sections of a rank-two bundle), and there should be an easy representation-theoretic way to see it... but I just cant't see it! |
Apr
9 |
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Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
... but my real question is another: why do we take the sum $g|_H+g^\sharp|_{H^*}$ and not, e.g., the difference of the two metrics on $H$ and $H^*$, in order to define the metric on $C_H$? And, most importantly, $\mathrm{id}_H\in\mathrm{End}(H)$ is also a metric on $C_H$. In other words, I suspect that there is more than one way to define a sub-Riemannian structure on $S^3\times\mathbb{RP}^2$, i.e., that we are in presence of a "multi-sub-Riemannian structure". So, my real curiosity is: how many $S^3$-invariant metrics exist on the contact distribution of $S^3\times\mathbb{RP}^2$? |
Apr
9 |
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Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
You're right - I've been sloppy about the construction of the metric on $C_H$. Nevertheless, the example I had in mind does not suffer from my sloppiness, since it deals with the case when $M$ is a Lie group, and more precisely the 3-sphere. In this case, $\mathbb{P}T^*S^3=S^3\times\mathbb{RP}^2$, and $C_H=H\oplus (H^\perp\otimes H^*)$, and then $S^2C_H^*=S^2H^*\oplus\mathrm{End}(H)\oplus (S^2H\oplus H^{\perp\ast\otimes 2})$ clearly contains the distinguished element $g|_H+g^\sharp|_{H^*}$ (the twisting factor $H^{\perp\ast\otimes 2}$ can be forgotten)... |
Apr
8 |
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Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian?
Thanks for the reference - though it'll be hard to dig out what I need. |
Apr
8 |
asked | Is $\mathbb{P}T^*M$ a sub-Riemannian manifold if $M$ is Riemannian? |
Apr
3 |
awarded | Yearling |
Mar
21 |
comment |
Invariant Lagrangians of a connection and its derivatives: how do they look like?
@WillieWong: ok, you answered my question! Believe it or not, I've asked this to several physicists - some of them with a long publication record in GR - and the answer was always the same "I'm almost certain it is true, but I cannot prove it, and I wouldn't know where to look"! |
Mar
21 |
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Invariant Lagrangians of a connection and its derivatives: how do they look like?
@WillieWong as $d^m\boldsymbol{x}$ I'm using an arbitrary volume form, just to identify horizontal $m$-forms on $J^n(C(M))$ with smooth functions. So: yes, my question is about natural invariants (if by "natural" one means $\mathrm{Diff}(M)$-invariant) of order $\leq n$ of torsion-free linear connections on $T M$. Now: what did you mean by "In some sense"? And why polynomials in the curvature should be enough? This way - I guess - you get only first-order invariants. What about the higher-order ones? Are they polynomial in the covariant derivatives of the curvature? |
Mar
21 |
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Invariant Lagrangians of a connection and its derivatives: how do they look like?
@RobertBryant sure, I've been sloppy. I used the name "Lagrangian", instead of "function", just because the question arose in the Lagrangian formulation of certain extended theories of gravity. Of course, my question is about the coefficient of $d^m\boldsymbol{x}$ in the correct expression $L d^m\boldsymbol{x}$ I should have used above. This is the typical "folkloristic question" appeared in mathematical physics: everyone claims it is true, but so far nobody was able to give me the slightest evidence of its validity. I'm sure an expert of differential invariants should know how to work it out. |
Mar
21 |
asked | Invariant Lagrangians of a connection and its derivatives: how do they look like? |
Mar
13 |
comment |
Contact distributions on $(G_2,P)$-type Cartan geometries in dimension 5
You gave a sophisticated answer to a lousy question: chapeau! |
Mar
13 |
accepted | Contact distributions on $(G_2,P)$-type Cartan geometries in dimension 5 |
Mar
8 |
comment |
Contact distributions on $(G_2,P)$-type Cartan geometries in dimension 5
Book-writing is a good entertainment for people over sixties - so let's discuss it in two decades - but a review paper why not? So, the two homogeneous 5-folds of $G_2$ are $S^2\times S^3$ and $S^5$, which are both contact, but only the former is contact homogeneous. The latter should be equipped with a $(2,3)$-type flag of distributions, but I fail to see it: is it evident? In fact, Sagershnig's paper says that the projectivised null cone in the imaginary octonions is $S^2\times S^3$, i.e., the opposite of what you say! That's why I'd like to see how a 2D and a 3D distribution on $S^5$ arise. |