liegroups

[liegroups]

（Still about hw14. ex2. d) Why m complementary to I+iI implies that m+ih\cap I+iI equals to ij\cap I+iI? We may write this more concretely, for arbitrary M\in m and iH\in ih, M+iH belongs to I+iI iff iH=A_1+iA_2-M for some A_1,A_2\in H, but is is not an obvious thing that A_1=0 since it seems not easy to exclude the case pr_{\mathbb{R}}(M)=A_1\in I\cap m_{\mathbb{R}}, this is what we want to show in e......

Okay, let me spell it out then. The easiest is probably to prove equation (4) directly, that is, show that m + i h = m \oplus i l.

Fix a complex-valued inner product on g, so that m is the orthogonal complement of l + i l. Decompose i h orthogonally as i l \oplus i a. Then i a is orthogonal to l + i l, so contained in m. Consequently m + i h = m + i a + i l = m + il = m \oplus i h.

[liegroups]

For hw14 ex2. d，why the isomorphism g/(m+ih)\cong I+iI/iI holds? I am a bit confused about showing the injectivity of the natural inclusion from I+iI/I to g/(m+ih), or equivalently. why (m+ih)\cap (I+iI) is contained in iI?

Consider the canonical map l + i l -> g / (m + i h). It is surjective because m is complementary to l + i l. Its kernel is the intersection of l + i l with m + i h, which equals the intersection of l + i l with i h, again because m is complementary to l + i l. As l is a subspace of h, the kernel of the above canonical map is i l.

[liegroups]

Will the exam involves contents of the differential geometry further than smooth manifolds, submanifolds, smooth maps, tangent spaces, vector fields, flow/integral curves of vector fields, differential forms, integration with differential forms, immersions and submersions, as described in the prequisites? For example, the Riemannian manifold, which occured in hw 8.

The exam will stick to the prerequisites as you list. Riemannian metrics etcetera were only there for the purpose of a particular exercise.

[liegroups]

Could I make sure is the exam and retake this year also open book? If so, could we bring anyrhing other than a copy of lecture notes and additional notes? For example the answer of homeworks and my own notes. Thanks very much

Indeed, the exam for Lie groups will be open book, like last years. You are allowed to use the lecture notes by Van den Ban and my additional notes, but no further texts,

[liegroups]

Hello, I had a question about question 4b for the homework. I have not been able to make sense of convergence. What topology are we considering on Aut(G)?

Exercise 4.b is only about pointwise convergence of the sequence f_n. For that no topology on Aut(G) is needed.

[liegroups]

Hello,I had a question about question 4b for the homework. I have not been able to make sense of convergence. What topology are we considering on Aut(G)?

Meant is: prove that the sequence f_n converges pointwise. So for each g \in G, you have to prove that lim f_n (g) exists in G. That determines a map f from G to G, and you have to show that f is a group automorphism of G. No topology on Aut(G) is involved here yet. 

[liegroups]

Good day prof, In the proof of lemma 22.5 (1), which give the matrix coefficientes of pi^* we have: m_{w,v}(g)=v(pi*(g)w)=w(pi(g^{-1})v)=.... why is the last equality true? If I use the definition of pi* i get this: m_{w,v}(g)=v(w\circ\pi(g^{-1})) but I don't know how to proceed from here. Thanks

Here v \in V, w \in V* and \pi is representation of G on V if I recall correctly. We can regard v also as an element of the linear dual space of V*. That means that v(w) equals w(v) be definition. Now plug the definition of \pi*(g) into 

v(\pi*(g)w) = (\pi*(g)w) (v).

[liegroups]

Good morning I have couples of doubts about the proof of Schuur lemma (2) ...1) when proving the surjectivity, when proving the surjectivity of T \in Hom_G(V,V')\{0} you used that dim(im T)\le min(dim V,dimV'), why is that? I agree that dim(im T)\le dimV' because im T is a subspace of V', but what why is dim(im T)\le dimV considering that V might (at this point of the proof) be infinite dimensional. 2) After we proof that T was bijective how do we prove that T has a continuous inverse?

For any linear map T between two vector spaces V and W one has dim (im T) \leq dim W because im T \subset W and \dim (im T) \leq \dim V by the dimension theorem from linear algebra. The latter has a meaning even if dim V is infinite, then it puts an upper bound on the cardinality of a basis of im T.   

By assumption V or V’ has finite dimension in this version of Schur’s lemma, so dim (im T) is bounded above by two cardinal number of which one is finite.

At the point in the proof where we know that T : V -> V’ is a linear bijection, we have also established that dim V = dim V’ is finite. Then T^{-1} : V’ -> V exists as a linear map, and it is continuous because every linear map between finite dimensional vector spaces is continuous.

[liegroups]

We have defined in lecture 11 the contragredient representation as \pi*(g)w=w\circ\pi(g^{-1}). Is this a standard definition? In every other place like in wikipedia or in Hall's book i find that the definition has a transpose: \pi*(g)=pi(g^{-1})^T while ours don't. However in lecture 12 you wrote when analizing the character of \pi* \pi*(g)=pi(g^{-1})^T which now agrees with Hall's book but now with our definition in lecture 11. How do I resolve this?

Please note that the definition of a representation also includes the underlying vector space. If \pi is a representation on a vector space V, then by definition \pi* is a representation on a vector space V* dual to V. In the lectures this was specified as the space of continuous linear functionals on V, but variations on that would also be possible. For the contragredient representation of \pi on V* there is only one possibility, namely \pi*(g)w = w \circ \pi(g)^{-1}. This is the definition of the contragredient representation in general.

In specific cases there are alternative definitions. Namely, if V is endowed with a nondegenerate bilinear form, then one can identify V with a linear dual space of V  via this form. (In the lecture I did this via the choice of a dual basis, when V has finite dimension.) Then the contragredient representation can be defined on V, with the formula \pi* (g) = \pi (g)^{-T}. Here the transpose is defined in terms of the bilinear form.

[liegroups]

Good morning Professor. In exercise VI.2b it is stated that "the R-action stabilizes the subset N = [0,+\infty) x (0,+\infty)". What does that mean? and how can I see that for the given action t(x_1,x_2)=(e^t x_1, e^{-t} x_2) for example?

Let us denote the action by \alpha. Then it means that \alpha (t,N) = N for all t in R. So the map "action of t" sends the set N to itself, for all t.

[liegroups]

Good evening Prof,

Does continuous action or smooth action withouth specifying if it's right or left means a) it is both i.e that G is abelian? b) left action? c) can't be any. I personally think is c), so in particular for exercises 1 and 2 of week 6 it should be enough to prove it for left actions and just state that the proof is symmetrical for right actions. Am I right?

In this case, it can be a left action or a right action, that is just not specified. But, as I explained in the lecture, one can easily pass from a right action (x,g) -> \beta (x,g) to left action, namely by defining \alpha (g,x) = \beta (x,g^{-1}). Therefore it suffices to solve this exercise for either a left action or a right, you may chose which.

[liegroups]

Dear Professor, Besides the lecture notes (section 13) which define a proper map as a map between Topological locally compact Hausdorff spaces, we have somethign similar in the book of Duistermaat (pag 53): " Recall that a continuous mapping \Phi from a topological space U to a topological space V is said to be proper if \Phi^{-1}(K) is compact in U for every compact subset K of V. (If U and V are Hausdorff, this implies that \Phi(C) is closed in V for every closed subset C of U.)" So according to the books it looks like we need Hausdorfness of both X and Y in f: X-->Y to prove that if f is proper then it is closed. I know it is saying it is a sufficient condition, but no other resource states that it is true withouth this assumption

Hmmm, I am not sure of this if X,Y are Hausdorff without further assumptions… Maybe Duistermaat implicitly assumed local compactness, I don’t know.

Anyway, my proof does not require X to be Hausdorff, assuming that Y is a metric space. I found this argument myself, but already on Wikipedia you can read that the statement of the exercise holds.

[liegroups]

Good afternoon Prof Are you absolutely sure that we don't need X and Y to be locally compact Hausdorf? I'd also like to point out that the def of proper map given in that question: "f is proper if f^{−1}(C) is compact for every compact subset C ⊂ Y" is different from the ones in the notes (section 13): "a continuous map f : X --> Y between locally compact Haus. (top.) spaces X and Y is proper if for every compact subset C in Y the preimage is compact. Did you purposely changed the def?

Good point, I had forgotten that in lectures notes properness is defined this way. The most general and most common definition of properness does not put restrictions on the kind of topological spaces. See for instance Wikipedia.

The statement of exercise VI.1.a is also true if Y is locally compact Hausdorff (and not necessarily metric). That is a better-known version, of which you can easily find a proof in the literature and on the internet. I felt it would be too easy to pose an exercise whose answer could simply be copied from a website, so I replaced it by the same question with Y metric instead, That statement is also known, but the argument requires more independent thinking.

[liegroups]

I found out that some people consider Hausdorf as part of the definition of compact, is this the case in exercise 1a of week 6.? From wikipedia: "Some branches of mathematics such as algebraic geometry, typically influenced by the French school of Bourbaki, use the term quasi-compact for the general notion, and reserve the term compact for topological spaces that are both Hausdorff and quasi-compact"

Indeed, some people see it that way. But most mathematicians do not, and in these exercises for Lie groups Hausdorff is not included in compactness. 

In fact, there are interesting examples of compact non-Hausdorff spaces, like the unit circle in C with two inseparable points 1′ and 1′’ instead of 1. Those are relevant in point-set topology and in fields like noncommutative geometry..

[liegroups]

Good evening, in exercise 3, part b of the last hw, Y is a left-invariant vector field, using the definition of left invariance, I have a bunch of tangent maps applied to Y. In order to do the computations, can I take Y as the derivative at 0 of exp(tY)?So, I am considering t maps to exp(tY) to be the integral curve of Y, although I not very sure since we only have seen it in the particular case that we were in some matrix group and exp (tY) coincided with the matrix exponential.

Yes, Y is the derivative of exp(tY) at  t=0. And indeed t -> exp (tY) is an integral curve, namely of the left invariant vector field v_Y given by v_Y (g) = T_e (l_g) Y. That is essentially the construction of the exponential map, as explained in the lecture.

[liegroups]

Do we have recording for Lecture 8 ?

How do you mean, lecture 8? We are only in the third week...

[liegroups]

Good morning, Professor . In exercise 3 of week2, they call $\Phi_X^t$ the flow of $v_X$ at time $t$. My question is isn't this the same as maximal integral curve $alpha_X(t)$ from the lecture ? I meant you just changed the notation? Besides that, this $\Phi_X^t$ has domain M, while our $alpha_X(t)$ has domain $t \in \mathbb R$ as any curve should, so it seems that they are not really the same. We haven't seen any equation relating them, How should get started with the solution of question 3a)?

Indeed, they are more or less the same. The difference is that for an integral curve one needs to specify where it starts. Suppose \gamma (t) is the integral curve starting at P for a vector field v. Then the flow of v satisfies \Phi (P,t) = \gamma (t).