Dear Prof. Tao,

Thank you very much for your attention. That was really useful.

Best wishes,

Arturo

is bounded by 1, supported in the precompact set , and is Borel measurable, hence in .

“ small enough” should really be “ small enough”, to ensure that lies in .

(Regarding the Notes 1 exercise, please post a detailed question in that post regarding the exercise, stating what you have tried and where precisely you are stuck.)

]]>Dear Prof. Tao,

Thank you very much. That was really useful.

Please, would you mind to help me again with my next questions?

First of all, in proposition 11, with the dilatation given by , to prove that is well defined we need to show that and are . Clearly, is simple so it is in . However, I do not see how to argue in the case of . Furthermore, I expect that it is continuous, but I do not know how to show that.

Secondly, in the proof of proposition 14, I do not understand the following sentence:

“Since , we see that does not lie in the -neighbourhood of if is small enough, where is the projection map.”

Why we get that? I do not see trivially a contradiction assuming that is contained in the closure of .

Also, in proposition 14, what about ? You express that it will be chosen later, but I do not see how and where we need to choose .

Finally, if it is possible, I would sincerely appreciate any hint to show the lower bound of point (vii) of exercise 9 of notes 1.

Again, thank you very much.

Best wishes,

Arturo

I guess strictly speaking one should not just use the fact that , but rather that for all . This is then incompatible with the escape property if for any , and since , this only leaves the possibility that .

]]>I have a question about the proof of theorem 8. There, to prove {\|g^y\|\ll\|g\|}, you say

“We can achieve this by the escape property (5). Let {n} be a natural number such that {n \|g\| \leq \epsilon}, then {\|g^n\| \leq \epsilon} and so {g^n \in B(0,\epsilon)}. Conjugating by {y}, this implies that {(g^y)^n \in B(0,5\epsilon)}, and so by (5), we have {\|g^y\| \ll \frac{1}{n}} (if {\epsilon} is small enough), and the claim follows.”

For me, it seems that you are using the escape property under the hypothesis that {\|g^n\|} is small. However, the escape property is given when {n\|g\|} is small.

Similarly, in Notes 2, proving that the one-parameter subgroups are Lipschitz, it also seems that you used the escape property when {\|\phi(\frac{t}{n})^n\|} is small, while it should be used when {n\|\phi(\frac{t}{n})\|} is small.

Of course, when {n\|g\|} is small, we have {\|g^n\|\leq n\|g\|}, concluding that {\|g^n\|} is small. However, I do not see a way to conclude that {n\|g\|} is small assuming that {\|g^n\|} is small. It is particularly annoying, since the escape property is actually giving that control.

In other words, my problem could be rewritten as follows:

Why is not possible to have a sequence {(g_k)_{k\in {\bf N}} } and a constant {C} such that {\|g^n_k\|\rightarrow 0} while {k\rightarrow \infty} but {n\|g_k\|>C} for each {k\in{\bf N}}?

Best wishes,

Arturo

Oops, I forgot to add the hypothesis that U (and hence ) was precompact, in which case one can make as small as desired by a compactness argument.

]]>Why can we choose V small enough, such that ? The fact that we don’t know a lot about worries me.(especially when I think about xy and )? Thank you.

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