User paul reynolds - MathOverflow

Answer by Paul Reynolds for Trichotomies in mathematics

2013-02-02T21:24:59Z

A simple yet useful one: An irreducible complex representation of a compact Lie group is either 'real', 'quaternionic', or 'complex'. That is, it is the complexification of a real irreducible, or it can be considered quaternionic through the existence of an equivariant conjugate-linear (real-)automorphism $j$ that squares to $-I$, or it is neither.

The statement combines Schur's Lemma and the fact that there are three associative real division algebras, here seen through complex eyes.

Answer by Paul Reynolds for Are properties of geodesics on a cylinder unique to cylinders?

2012-07-07T22:33:52Z

I wanted to show that this property of geodesics doesn't even characterise the topology of the surface, and I think this is an example.

This is a bunch of cylinders of the kind Rbega mentions, deformed slightly so as to be joined smoothly together. The cut off edges should be asymptotic to parallel planes.

Answer by Paul Reynolds for n-dimensional "cross product" reference request

2012-04-18T11:29:20Z

Here's a bit of history on cross products that, if not directly useful, will hopefully provide some context. They were defined in "Beno Eckmann, Stetige Losungen linearer Gleichungssysteme, Comment. Math. Helv. 15(1943)" as follows: An $r$-fold cross product on a real vector space $V$ of dimension $n$ with inner product $g$ is a continuous map \[ P : \underbrace{V \times \cdots \times V}_r \to V \] satisfying

1) $P$ is skew;
\[ g\big( P(v_1, \ldots ,v_r),v_i\big) = 0 \ , \ 1 \leq i \leq r \ , \] 2) $P$ respects $g$;
\[ g\big(P(v_1, \ldots ,v_r),P(v_1, \ldots ,v_r)\big) = \det g(v_i,v_j) \ . \]

These were classified by Eckmann and Whitehead (see "George W. Whitehead, Note on cross-sections in Stiefel manifolds, Comment. Math. Helv.37 (1962/1963)") using algebro-topological methods. They were later also classified by Brown and Gray (see "Robert B. Brown and Alfred Gray, Vector cross products, Comment. Math. Helv. 42 (1967)") where those authors included an extra axiom: $P$ has to be multilinear. This extra axiom makes no significant difference to the classification. The classification theorem is:

An $r$-fold cross product on a real vector space $V^n$ exists if and only if we have one of

$\bullet$ $n$ even, $r=1$,
$\bullet$ $n=7$, $r=2$,
$\bullet$ $n=8$, $r=3$,
$\bullet$ $n$ arbitrary, $r=n-1$.

The proof of Brown and Gray uses Hurwitz' structure theorem for composition algebras. If you add a dimension to $V$ you can define a composition algebra, and vice versa. Their paper is my favourite reference. They actually consider a more general situation where the bilinear form is indefinite, that leads to more cross products (but only in the four cases above). They even work with any field of characteristic not $2$.

Using the standard inner product on $V = \mathbb{R}^n$ (which you are implicitly using by referring to $\ast$), your cross product is the last one on the list (it is a cross product in this sense, I was wrong to say otherwise earlier). It is normally just called the volume form, with the appropriate identifications made. I don't know about the earlier history of that particular case, or whether this was thought of as a cross product prior to the papers I've mentioned.

As Ryan points out, in two dimensions there is a $1$-fold cross product, and this is rotation by $90^{\circ}$. That's because $1$-fold cross products are the same as complex structures. The usual $2$-fold cross product on $\mathbb{R}^3$ is the volume form, and fits into the fourth case of the classification.

I cannot miss the chance to briefly mention the role of cross products in geometry. One can define a cross product on the tangent bundle of a Riemannian manifold. Kaehler manifolds are those with (parallel) $1$-fold cross products, and the cross products in seven and eight dimensions correspond to the exceptional holonomies.

Answer by Paul Reynolds for Holonomy group of cotangent bundle

2012-03-19T18:56:58Z

No. Of course, we must first assume $\mathcal{M}$ is non-flat to have any chance. Then, while it is true that the Sasaki metric on the tangent bundle $T\mathcal{M}$ along with the canonical symplectic structure form an almost-Hermitian structure, its torsion need not vanish. Even when it does, the holonomy group may not equal $SU_n$; the tangent bundles of complex projective spaces are hyper-Kaehler.

Comment by Paul Reynolds

2013-04-09T19:11:54Z

This condition is easy to say if you think of the Levi-Civita connection as an Ehresmann connection, i.e. as a distribution on $F$. Then, $P$ being parallel just means that the distribution is tangent to it.

Comment by Paul Reynolds

2013-02-23T19:12:34Z

There are lots of nice explanations of torsion here mathoverflow.net/questions/20493/… although you may consider them to be advanced also.

Comment by Paul Reynolds

2013-02-17T15:48:52Z

It seems there is an orthogonal/skew-symmetric mix-up in the wording of this question, possibly from $SO_n$/$\mathfrak{so}_n$.

Comment by Paul Reynolds

2013-02-03T14:12:11Z

@Sam, in the setting of my answer, if you integrate $\chi(g^2)$ over all $g$ in the group $G$ where $\chi$ is the character of the rep, you get $1,0,-1$ when the rep is 'real', 'complex' or 'quaternionic' respectively. Another manifestation of the $1,0,-1$ trichotomy is in the behaviour of tensor products over $\mathbb{C}$ of the different types.

Comment by Paul Reynolds

2013-02-03T12:50:42Z

The curvature trichotomy indeed comes up in many different ways, e.g. look at the striking differences between positive, $0$ (hyper-Kaehler) and negative quaternionic-Kaehler geometry.

Comment by Paul Reynolds

2013-02-03T11:37:54Z

I'm glad Arnol'd puts Pontryagin classes in the quaternionic column. This point seems to often go unmentioned.

Comment by Paul Reynolds

2012-12-16T21:24:28Z

I thought the gauge group was the space of sections of $M \times G$. $P \times_G G$ is naturally isomorphic to $P$ unless I've misunderstood.

Comment by Paul Reynolds

2012-11-23T21:47:31Z

Thanks very much Richard! I will take the time to read this as it's something I've wanted to know for a while.

Comment by Paul Reynolds

2012-11-23T20:02:56Z

Interesting! I wonder how this varies with the fraction $r = 2/3$...

Comment by Paul Reynolds

2012-11-23T19:43:27Z

A agree that choosing $1$ is the only rational choice, but still I would like to actually do this experiment with a bunch of non-mathematicians and see what $2/3$ of the average turns out to be. I suspect it changes after the first go, or if you give them much time to think, but such empirical information could have been useful in a pub quiz I used to go to.

Comment by Paul Reynolds

2012-11-15T15:33:21Z

Given that you've accepted Robert Bryant's answer below, I think the answer to my question should actually be no. Unless I've misunderstood completely, your question is answered by the notion of holonomy subbundle.

Comment by Paul Reynolds

2012-11-14T18:37:22Z

Are you talking about the holonomy group of the total space of $OM$ with its usual (Sasaki-Moek/O'Neill) Riemannian metric?

Comment by Paul Reynolds

2012-11-06T16:05:28Z

$\mathbb{R}^8$ admits an $S^6$'s worth of compatible complex structures, given by multiplication by unit imaginary octonions. Do you want to add more restrictions?

Comment by Paul Reynolds

2012-10-26T17:36:38Z

In that case if you can find $g$ and $\mu$ then you can find them so that $\lambda = 1$. I was going to say something about the case $X$ is non-vanishing, but it's pretty much what is said in the paper you linked to.

Comment by Paul Reynolds

2012-10-25T23:35:50Z

Do you require that the volume form be that given by the metric (and orientation), or is it chosen separately? I'm guessing the former but your notation makes me unsure.