User eric o. korman - MathOverflow

Recovering a Lie algebra from its affine Lie algebra

2012-05-24T14:59:13Z

For a complex simple Lie algebra $\mathfrak g$ let $\hat{\mathfrak g}$ be its affine Lie algebra (see e.g. http://en.wikipedia.org/wiki/Affine_Lie_algebra#Definition for the definition). Is there an intrinsic way of recovering $\mathfrak g$ from $\hat{\mathfrak g}$? In other words, if I'm given an arbitrarily defined infinite-dimensional Lie algebra $\mathfrak h$ and I want to know if it's isomorphic to $\hat{\mathfrak g}$ for some $\mathfrak g$, is there some deterministic method of finding what $\mathfrak g$ would have to be?

Thanks!

Reconciling two notions of geometric quantization.

2013-04-11T17:50:57Z

Let $(M,\omega)$ be a compact symplectic manifold and $(L,\nabla)$ a prequantum line bundle. There are two schemes to quantize this data:

Choose a polarization $P$ of $M$ and define the quantum Hilbert space to be sections of $L$ that are parallel along $P$. This space admits an action of the Poisson algebra of smooth functions on $M$.
Choose a compatible almost complex structure. This gives a $Spin^c$ structure and define the quantum space to be the index of the corresponding Dirac operator twisted by $L$. Here there is no action of $C^\infty(M)$. When a Lie group $G$ acts on everything, the index is an element of the representation ring of $G$.

In the case that $M$ is Kähler, $L$ a holomorphic line bundle, and $P = T^{0,1}M$, the first method gives the space of holomorphic sections of $L$ and the second method gives the index of $\bar\partial_L + \bar\partial_L^*$.

The first method is more tied to physics and may seem a little ad-hoc from a mathematical point of view. The second seems more natural mathematically (since it fits in well with symplectic reduction and the index of an elliptic operator is more well-behaved mathematically than the kernel of $\nabla$ along $P$). But it seems to be a very weak notion of quantization since there is no action of $C^\infty(M)$. .

How can these two viewpoints be reconciled? Should the second version be viewed as just a more natural mathematical construction? Is there a nice way to tie it back to the physics?

Parallel forms and cohomology of symmetric spaces

2013-04-08T21:36:51Z

Let $G/H$ be a compact symmetric space. Then I believe the following is true: if $\alpha \in \Omega^k(G/H)$ and $\nabla$ the Levi-Civita connection, then $$ (\alpha \text{ is induced by an $\operatorname{Ad} H$ element of $\Lambda^* \mathfrak h^\perp$}) \Leftrightarrow \alpha \text{ is $G$-invariant} \Leftrightarrow \nabla \alpha = 0 \Leftrightarrow \alpha \text{ is harmonic}. $$ This means the real cohomology of $G/H$ is isomorphic to the space of $\operatorname{Ad} H$ invariant elements of $\Lambda^* \mathfrak h^\perp$, which seems like a reasonable thing to be able to compute.

For what compact symmetric spaces have these been computed explicitly, and where can this sort of thing be found in the literature?

Stiefel–Whitney classes in the spirit of Chern-Weil

2011-01-17T17:55:45Z

Chern-Weil theory gives characteristic classes (e.g. Chern class, Euler class, Pontryagin) of a vector bundle in terms of polynomials in the curvature form of an arbitrary connection. There seems to be no hope in getting Stiefel-Whitney classes from this method since Chern-Weil gives cohomology classes with real coefficients while Stiefel-Whitney classes have $\mathbb Z/2$ coefficients. Further, since any vector bundle over a curve has vanishing curvature, classes obtained by Chern-Weil can't distinguish, for example, the Mobius bundle from the trivial bundle over the circle (while Stiefel-Whitney classes do).

Nonetheless, I am wondering if there is a more general or abstract framework that allows one to define the Stiefel-Whitney classes in the spirit of Chern-Weil. For example, maybe this is done through a more abstract definition of a connection/curvature.

Parallel orthogonal complex structures on complexified tangent bundle.

2012-12-06T21:26:14Z

Let $M$ be a Riemannian manifold. I'm wondering if there are nice conditions/obstructions or (non-Kahler) examples of the existence of a map $J \in \operatorname{End} (TM \otimes \mathbb C)$ such that

$J^2 = -Id$.
$g(JX,JY) = g(X,Y)$ for all $X,Y \in TM \otimes\mathbb C$.
$\nabla J = 0$ where $\nabla$ is the Levi-Civita connection.

Of course, if $J$ preserves $TM$ then $M$ is Kahler which is a pretty strong condition. But, for example, if $J$ does not preserve $TM$ then the 2-form $\omega(\cdot,\cdot) = g(J\cdot, \cdot)$ is closed and non-degenerate but could now be zero in de Rham cohomology since it's not necessarily real.

Examples of Clifford Modules

2012-12-03T16:09:28Z

For a Riemannian manifold $M$, a Clifford module is a bundle over $M$ that is, fiberwise, a representation of the Clifford algebra bundle $Cl(TM)$ and has a connection compatible with this action. The canonical examples of Clifford modules are

$\Lambda^* M$ (in two ways, with the grading formed by the degree of differential forms or with the hodge star).
The spinor bundle associated to a spin or spin$^c$ structure.
The Dolbeault complex $\Lambda^{0,*} TM$ for a complex manifold (which is really a special case of a spin$^c$ structure).

Are there other examples of Clifford modules that come up naturally?

Summation methods for divergent series

2010-03-24T15:16:49Z

There are many methods for assigning a value to a series that diverges, e.g. zeta function regularization, Abel summation, Cesaro summation, etc. From all of the examples I've found, two methods either give the same result or one of them doesn't work. For example, both zeta function regularization, Ramanujun summation, and a method of Euler assign -1/12 to 1 + 2 + 3 + 4 + ... while Abel summation can't assign a value. My question is if there is an example of a series that different summation methods assign different values to or is it the case that any two summation methods must agree on divergent series (that they can assign a value to). Here I am assuming that both summation methods assign the correct value to convergent series and are linear. I am guessing we need stronger conditions since it seems that the space of convergent series is not dense, in some sense, in the space of all series.

EDIT: I was able to pick up a copy of Hardy's "Divergent Series." It's a really neat book but I have yet to be able to find in it an example of a divergent series that gets assigned two different values by two different linear and consistent summation methods. He does show how a method being linear forces specific series to have a unique value. Surely the issue of whether two general summation methods (with reasonable conditions that they satisfy) can disagree on a certain series must come up somewhere in the literature.

Index of a differential operator between trivial bundles.

2012-11-16T02:32:29Z

Let $M$ be a closed parallelizable manifold and $D: \Gamma(E) \to \Gamma(F)$ an elliptic differential operator between trivial vector bundles $E,F \to M$. The Atiyah Singer index theorem implies that the index of $D$ is zero. Is there a way to prove this with less machinery?

By the way, this question is a cross-post from math.SE.

EDIT:

Actually I think I made a mistake in my reasoning, which was that the symbol class $[\sigma_D] \in K(TM)$ is zero (I actually didn't need triviality of $TM$). Viewing $K$-theory as sequences of bundles the symbol class is $$ 0 \to \pi^* E \stackrel{\sigma_D}{\to} \pi^* F\to 0 $$ where $\pi: TM \to M$. Now if $TM^+$ is the one-point compactification of $TM$ then the isomorphism $K(TM) \to \tilde K(TM^+)$ is given by extending the sequence to $TM^+$. I thought that the extension would have to involve trivial bundles as well, from which it will follow that $\sigma_D = 0$ since for a compact space any sequences involving trivial bundles is zero in $\tilde K$. But now I think this extension need not involve trivial bundles: $K(\mathbb R^2) \simeq \tilde K(S^2) = \mathbb Z$. But every bundle over $\mathbb R^2$ is trivial so my argument would give $K(\mathbb R^2) = 0$.

Representation of double cover of $U(n)$ on eigenspaces of harmonic oscillator

2012-04-19T14:44:00Z

Consider the metaplectic representation of $Mp(n)$ on $L^2(\mathbb R^n)$. We can view $U(n)$ as a subgroup of $Sp(n)$ and so inside $Mp(n)$ is a double cover $\tilde U(n)$ of $U(n)$. The restriction of the metaplectic representation to $\tilde U(n)$ commutes with the Hamiltonian of the harmonic oscillator: $H = \sum_i (x_i^2 - \frac{\partial^2}{\partial x_i^2})$ and so decomposes as a direct sum of finite dimensional representations of $\tilde U(n)$ on the eigenspaces of $H$.

I am looking for a reference that discusses these representations of $\tilde U(n)$. Specific things I would like to know are if each of these representations is irreducible and if any of them descend to representations of $U(n)$.

Answer by Eric O. Korman for Representation of double cover of $U(n)$ on eigenspaces of harmonic oscillator

2012-04-22T15:48:56Z

I just found that Borel and Wallach's * Continuous cohomology, discrete subgroups, and representations of reductive groups* does a detailed analysis of this representation (which they call the oscillatory representation). It's on google books: http://books.google.com/books?id=_EZY9LhAxosC&lpg=PA257&ots=NIQffrLQCA&dq=borel%20wallach%20cohomology%20lie&pg=PA151#v=onepage&q&f=false

Answer by Eric O. Korman for Covariant derivative

2012-01-05T06:08:22Z

If $E \to X$ is a (finite-dimensional) vector bundle and $P$ is the principal $GL(n)$ bundle of frames, then there is a one-to-one correspondence between covariant derivatives on $E$ and principal connections on $P$. A good reference for details is proposition 4.4 of Lawson and Michelson's ``Spin Geometry" (unfortunately the relevant part is not on google books).

If $U$ is an open set of $X$ on which $E$ is trivial then relative to some local frame over $U$ we have the connection one-forms in $\Omega^1(U; \mathfrak{gl}(n))$. You can pull these back to $U \times GL(n)$, which then determines a $\mathfrak{gl}(n)$-valued one-form on $P\vert_U$ since the frame gives an isomorphism $P\vert_U \simeq U \times GL(n)$. Then one can show that these locally defined forms on $P$ piece together to form a global connection form.

Is an identity that is true for matrix Lie groups true for all Lie groups?

2011-11-22T14:27:53Z

Many identities for Lie groups are more easily proved for matrix groups. A non-trivial example is the equation $$ \frac{d}{dt}\vert_{t=0} \exp(-X)\exp(X+tY) = \frac{1-e^{-\operatorname{ad} X}}{\operatorname{ad} X} Y. $$

My question is if it is always sufficient to prove identities in the case of matrix algebras. For example, is there some sort of density argument using a topology on the space of Lie groups that makes matrix groups dense? Or perhaps there is an analogue of the notion of "permanence of identities" used to prove statements about matrices.

EDIT: Admittedly my question is vague, and as I responded to a comment, I am really looking for a heuristic that works in a large set of ``natural" cases.

Here is a good example of what I'm after is this mathoverflow question about showing if $\alpha$ and $\beta$ are one parameter subgroups of a Lie group then $(\alpha \beta)'(0) = \alpha'(0) + \beta'(0)$. This is very easy to prove for matrix groups since the product rule carries over. It is not hard to prove for general Lie groups (see e.g. my answer to that question) but it is more difficult in my opinion.

Complex structure on flag manifolds

2011-09-07T21:05:21Z

Let $G$ be a compact Lie group and $T$ a maximal torus of $G$. Then the flag manifold $G/T$ is a complex manifold and a symplectic manifold. One way to see the symplectic structure is to view $G/T$ as the co-adjoint orbit of a generic element $F_0 \in Lie(T)^*$. Then the symplectic structure is given by $$ \omega_F(X^+_F, Y^+_F) = F([X,Y]), $$ where $X^+,Y^+$ are the fundamental vector fields corresponding to $X,Y \in Lie(G)$ and $F \in Orbit(F_0)$.

All the references I've seen get the complex structure on $G/T$ by showing it is isomorphic to $G^{\mathbb C}/B$ where $G^{\mathbb C}$ is the complexification of $G$ and $B$ is a Borel subalgebra.

My question is if there is a way to get the complex structure explicitly in terms of the Lie algebras of $G$ and $T$, in a similar vein to how I defined the symplectic structure.

Examples of "folk theorems"

2010-07-18T21:48:09Z

In this post, Justin gives a quote about Raoul Bott that has this line in it:

He talked about 'folk' theorems... theorems everyone knew, but were never written down.

What are some good/interesting examples of these types of theorems?

Answer by Eric O. Korman for when can we lift an action of Lie algebra?

2011-08-24T14:58:04Z

Yes if $G$ is connected and simply connected, since in that case there is a one to one correspondence between Lie group homomorphisms $G\to H$ and Lie algebra homomorphisms $\mathfrak g \to \mathfrak h$. Since a representation of $\mathfrak g$ is just a Lie algebra homomorphism $\mathfrak g \to \mathfrak{gl}(V)$, your desired result follows.

EDIT: I'm assuming $V$ is finite dimensional.

Is there an intrinsic definition of the topological index map in $K$-theory?

2011-05-12T21:44:04Z

In the language of $K$-theory, the Atiyah-Singer index theorem says that for a compact manifold $X$ the topological index map $\text{t-index}: K(TX) \to K(T\mathbb R^n) \simeq \mathbb Z$ induced by embedding $X$ in $\mathbb R^n$ is equal to the analytical index map $K(TX) \to \mathbb Z$ obtained by looking at the index of the elliptic operator whose symbol corresponds to the given element in $K(TX)$.

My question is if there is a definition of the topological index map that does not require an embedding into a euclidean space. Clearly by the index theorem we can take the analytic index as a definition, but is there a more topological/geometric intrinsic definition for t-index that is (relatively) easily seen to be equivalent to the above definition?

Answer by Eric O. Korman for Lie group operation and tangent vectors

2011-04-25T11:15:56Z

Heres a pretty clean proof. Let $m:G \times G \to G$ denote the multiplication map. Then we have (identifying $T_{e,e} G\times G$ with $T_e G \oplus T_e G$) $$ m_*(\alpha'(0), 0) = \frac{d}{dt}\vert_{t=0} m(\alpha, e) = \alpha'(0). $$ The same thing shows that $m*(0,\beta'(0)) = \beta'(0)$. By linearity we get $m_*(\alpha'(0),\beta'(0)) = \alpha'(0) + \beta'(0)$.

Defining the inverse of the Bott map

2011-04-14T14:45:58Z

I hope this isn't too narrowly focused. I have a question concerning the inverse of the Bott map as defined in Atiyah's paper, Bott Periodicity and the Index of Elliptic Operators. On page 122 he defines it as the composition

$$K^{-2}(X) \to K(S^2 \times X) \stackrel{\text{index} \bar\partial}{\longrightarrow} K(X).$$

My question is: what is the first map? At first I thought it was $$ K^{-2}(X) = \tilde K(S^2 \wedge X^+) \to \tilde K(S^2 \times X^+) \to \tilde K(S^2\times X) \subset K(S^2 \times X) $$ where the first map is the pullback of the projection and the second is the pullback of the inclusion. But from the 6 term exact sequence, for this first map to be injective (which it needs to be for this to be correct) we need $$ K(S^2 \times X^+, S^2 \times X) \to \tilde K(S^2 \times X^+) $$ to be the zero map which doesn't seem correct since $S^2 \times X^+/S^2 \times X$ is identifiable with $S^2 \sqcup {pt}$ and if we pullback a non-trivial bundle over this space to $S^2 \times X^+$ it seems like it won't trivialize.

Thanks!

Answer by Eric O. Korman for Defining the inverse of the Bott map

2011-04-16T21:06:45Z

Actually I think I found a better way to look at the map, which shows that it is injective. Consider the pair $(S^2 \times X, {pt} \times X)$. Then since ${pt} \times X$ is a retract of $S^2 \times X$, the 6-term sequence in $K$-theory splits to give a short exact sequence $$ 0 \to K(S^2 \times X, {pt} \times X) \to K(S^2 \times X) \to K({pt} \times X). $$ But $(S^2 \times X) / (pt \times X) \simeq S^2 \wedge X^+$, giving the desired map.

Answer by Eric O. Korman for Conformal-symplectic geometry ?

2011-01-28T22:31:05Z

If the manifold has dimension bigger than 2, I think the conformal class of $\omega$ is just $k\omega$ for constants $k$. Locally, by Darboux we can write $\omega = \sum_i dq^i \wedge dp_i$. If the dimension is greater than 2, the only way for $0 = d(f\omega) = df \wedge \omega$ is if $df = 0$.

Which continuous functions are polynomials?

2010-08-03T18:21:08Z

I posted this on the new math.SE website but didn't get much of a response, so I am reposting it here.

Suppose $f$ is a continuous $\mathbb{R}$-valued function on $\mathbb{R}^n$. What type of conditions on $f$ guarantee it is a polynomial up to homeomorphism. That is, when can I find a homeomorphism $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $\phi^* f = f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?

Some related questions:

A necessary condition in the case of $n = 1$ is that point inverse images of $f$ must be finite (since a polynomial has only finitely many roots). Is this sufficient?
What if we replace $\mathbb{R}$ by $\mathbb{C}$?
What if we look at smooth functions and diffeomorphism instead? (I tried playing around with the implicit function theorem but didn't get anywhere).
What about the complex analytic case?

I'm not quite sure how to tag this, so feel free to edit them.

Answer by Eric O. Korman for Why is the x->x^1/3 atlas on R diffeomorphic with the x->x atlas on R?

2010-07-29T16:39:48Z

In regards to your first question, the last three statements are essentially the same since to talk about diffeomorphisms between two manifolds you need to have a smooth structure on those manifolds, which is a choice of an (equivalence class of) atlas. One would usually simply say that the manifold $R$ with its regular smooth structure is diffeomorphic to the manifold $R$ with the smooth structure determined by the atlas $x \mapsto x^{1/3}$.

Here, the different smooth structure on $R$ is provided by an atlas that is a homeomorphism $F: R \to R$. This will always produce a manifold diffeomorphic to $R$ with the regular smooth structure since a diffeomorphism between them will be $F^{-1}: R \to R'$, where $R'$ is $R$ with the atlas $(R, F)$.

I do not know a lot about exotic structures, but I do not believe that there are examples where it is obvious that two smooth manifolds, that have the same underlying topological manifold, are not diffeomorphic. For example, I believe Milnor proved that some of his exotic 7-spheres were not diffeomorphic using Morse theory.

Answer by Eric O. Korman for Lie group actions and f-relatedness

2010-07-13T00:48:56Z

Since I could use the reputation... letting $\psi: G \to \operatorname{Diff}(M)$ be the action then $X' = \psi_* X$, where we identify $\operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $\operatorname{Diff}(M)$. Then $[X', Y'] = [X,Y]'$ is immediate.

Answer by Eric O. Korman for Demystifying complex numbers

2010-07-02T15:12:42Z

Tristan Needham's book Visual Complex Analysis is full of these sorts of gems. One of my favorites is the proof using complex numbers that if you put squares on the sides of a quadralateral, the lines connecting opposite centers will be perpendicular and of the same length. After proving this with complex numbers, he outlines a proof without them that is much longer.

The relevant pages are on Google books: http://books.google.com/books?id=ogz5FjmiqlQC&lpg=PP1&dq=visual%20complex%20analysis&pg=PA16#v=onepage&q&f=false

Answer by Eric O. Korman for Functional Analysis and its relation to mechanics

2010-07-01T00:23:37Z

One of the biggest problems in mathematical physics is actually to understand the link between Hamiltonian/Lagrangian mechanics and functional analysis. This is because classical mechanics is formulated in the former setting while quantum mechanics is formulated in the functional analysis setting. The act of going from classical mechanics to quantum mechanics is called quantization and basically consists of assigning functional analytic operators to classical observables, in a way that respects the Poisson and Lie brackets. For example in classical quantization we assign position to the operator of multiplication by x and we assign to momentum the operator $-i\frac{d}{dx}$. Both of these act on (a dense subset of) the space $L^2(\mathbb R)$, which is taken to be the space of wave functions in one dimension. You may want to take a look at the orbit method, which is the mathematics involved in a quantization scheme called geometric quantization.

Some relevant MO discussion about this are:

http://mathoverflow.net/questions/6200/what-is-to-quantize-something

http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean

Answer by Eric O. Korman for What are the most elegant proofs that you have learned from MO?

2010-06-27T02:06:37Z

Unfortunately I can't find the link but someone mentioned this proof that there are irrational numbers $a$ and $b$ such that $a^b$ is rational: if $\sqrt{2}^\sqrt{2}$ is rational then we are done, if it is irrational then $2 = (\sqrt{2}^\sqrt{2})^\sqrt{2}$ is an irrational raised to an irrational.

When is a conjugacy class of matrices an embedded submanifold?

2010-05-30T00:45:46Z

Let $M_{n\times n}$ denote the set of $n\times n$ real matrices and let $GL_n$ be the subgroup of invertible matrices. $GL_n$ acts on $M_{n\times n}$ smoothly by conjugation, which means that each conjugacy class (which is an orbit of this action) is an immersed submanifold of $M_{n\times n}$. However, the action is not proper (e.g. the isotropy groups are not compact) so the orbits may not be embedded submanifolds.

My question is if there are nice conditions on a matrix that guarantee that its conjugacy class is or is not an embedded submanifold. My interest in this question actually comes from trying to understand the space of all complex structures on a real vector space: it can be shown that the set of all complex structures is the conjugacy class of the block matrix $\begin{pmatrix} 0 & -I \\ I & 0\end{pmatrix}$ and I was wondering if this is an embedded submanifold. So an answer to this question (if the above doesn't have a nice answer) would also be appreciated.

Answer by Eric O. Korman for Representations of Pin vs. Representations of Clifford

2010-05-11T01:15:31Z

I think this argument works in the case that the inner product is definite (which is the case you are considering anyways). In this case, the Pin group is generated by the set of all normalized non-zero vectors. Therefore, by the universal property of Cl, a rep of Pin gives a unique rep of Cl and restricting this rep gives you back the rep of Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a rep of Cl. If you know where all the $e_i$'s get mapped, you then get a unique rep of Cl which gives a unique rep of Pin.

If the inner product is not definite you will still get a rep of Cl but I can't see how you know that it will agree with the original rep of Pin.

EDIT: I have assumed that the map we get from the set of all unit vectors (by restricting the map from Pin), is the restriction of a linear map from $R^n$. This doesn't seem to be obviously the case, but is equivalent to the statement that every rep of Pin is the restriction of a rep of Cl.

Answer by Eric O. Korman for Is there any geometry where the triangle inquality fails?

2010-05-01T12:30:41Z

I just wanted to add that in the Lorentzian case, the triangle inequality gets reversed for a certain class of vectors (those with positive time component). Since the norm of a vector corresponds to the elapsed time measured by a clock moving along the vector, a physical consequence of this is the twin's paradox: by flying out in a spaceship from earth and then returning, you age less than if you had just stayed still.

Clifford Algebra in Dirac Equation

2010-03-14T05:05:27Z

I am wondering if there is any mathematical (or physical, besides the fact that classical quantum mechanics uses complex numbers) justification for why the complexified (1,3) Clifford algebra is used in Dirac's equation. A (the?) key point of special relativity is that spacetime is a real 4-d vector space with an inner product of signature (1,3). But by complexifying the signature becomes irrelevant-- all complex Clifford algebras in a given dimension are isomorphic where for real Clifford algebras, even signatures (p,q) and (q,p) are not isomorphic in general (as a side question: can there be a physical significance to this fact? or do only the spin group and the even subalgebra of the Clifford algebra, which are the same for (p,q) and (q,p), matter? I only hear about spinor bundles and spin structures, never pinor bundles or pin structures).

Thanks and I hope this isn't too physicsy of a question!

Comment by Eric O. Korman

2013-04-12T02:27:38Z

@Urs Schreiber: Thanks for your comment. I've skimmed through some of Landsman's work and it does seem to incorporate both notions of quantization. Indeed, it seems that this question is one of his motivating factors.

Comment by Eric O. Korman

2013-04-09T00:03:36Z

Thanks for the answer. I think I do mean the Levi-Civita connection and not a Laplacian. At least for $G$ compact, isn't any harmonic form necessarily $G$ invariant and for a symmetric space any invariant form is parallel. I will look into using that Maple package, though I'm still wondering if there's a list out describing $\Lambda^* \mathfrak h^\perp)^H$ for a large class of examples.

Comment by Eric O. Korman

2012-12-07T03:19:25Z

Thanks for the great answer.

Comment by Eric O. Korman

2012-12-03T23:23:03Z

@Robert Bryant: I did but didn't find any examples beyond the ones I gave.

Comment by Eric O. Korman

2012-11-16T19:51:55Z

Thanks for the nice example!

Comment by Eric O. Korman

2012-11-16T14:10:29Z

@Johannes, Paul: Now I think I was actually wrong-- I edited the question with my reasoning and will probably delete it soon.

Comment by Eric O. Korman

2012-05-24T16:33:03Z

Sorry my question was worded kind of vaguely. Instead of saying "I know it's isomorphic to..." I really should have said "I want to see if its isomorphic to...then what would $\mathfrak g$ have to be". I have just edited it.

Comment by Eric O. Korman

2012-04-22T15:49:45Z

Thanks for the response. It turns out that the reps do not descend to $U(n)$ in general but do to $SU(n)$.

Comment by Eric O. Korman

2012-04-22T15:46:06Z

Thanks for the answer. It may take me some time to digest it.

Comment by Eric O. Korman

2011-11-23T03:12:01Z

@Mariano: no, I don't. I just gave that example to illustrate that in proving a statement about a Lie group, I don't see how one can reduce the result to a locally isomorphic group. I think for the type of identities I'm thinking about, it will be sufficient to prove them for the universal cover. But the only examples of non-matrix Lie groups I know of are simply connected (the universal covers of $SL(n,\mathbb R))$.

Comment by Eric O. Korman

2011-11-23T00:58:52Z

@Alain: is the identity in my question really just at the Lie algebra level? The left hand side involves the pushforward of multiplication by $\exp(X)$ at the point $\exp(X)$, which would seem to depend on the group.

Comment by Eric O. Korman

2011-11-22T23:30:19Z

@Mariano: you're right and I agree my question is a little vague. I am really looking for a heuristic that works in many natural cases, in a similar vein to how in proving identities involving $\tr$ and $\det$, one may assume that the matrix involved is diagonalizable.

Comment by Eric O. Korman

2011-11-22T19:45:48Z

i dont see why local iso is sufficient if the identity is at the group level. e.g. in S1 there exists nonzero X st exp(X+Y)=expY for all Y. this doesnt hold for R

Comment by Eric O. Korman

2011-09-08T02:50:17Z

Thanks for the answer. I'm just a little confused about your last paragraph. Is it saying that any almost complex structure on $G/T$ is integrable?

Comment by Eric O. Korman

2011-08-03T13:49:55Z

@J. Fabian: you're right-- my $h$ depends on $x$. I don't see how to get rid of this dependence.