20,348 reputation
13058
bio website facebook.com/paterz1
location Montreal, Canada
age 23
visits member for 3 years, 3 months
seen 2 hours ago

I finished my undergraduate studies in mathematics at University of Montreal. I am doing my Ph.D at the Berlin Mathematical School in algebraic geometry. My current interests, besides algebraic geometry, are number theory, analysis, topology, measure theory, representation theory and commutative algebra.


4h
comment Is every finite group of isometries a subgroup of a finite reflection group?
You might be interested in this : en.wikipedia.org/wiki/…
5h
comment When does L' Hopital's rule fail?
@Bombyxmori : I'm not talking to a student, am I? :) You have much more polite ways of saying this, such as "Try a different approach, you can do it!" or "Have you tried to =*insert different approach here*=?". I was tired, I agree I was probably way too rude to write an answer on MSE...
14h
revised Basic question about conditional expectation
corrected typos
14h
comment When does L' Hopital's rule fail?
@Bombyxmori : JimmyK4542 gave us examples where thinking is the best way to go ; I cannot imagine how extremely complicated such a general would be if it could exist. Using his/her own brain should be the most satisfactory thing to a student.
23h
comment When does L' Hopital's rule fail?
+1, good examples. I was trying to come up with such an example (and I utterly failed at it..).
23h
comment When does L' Hopital's rule fail?
@Bombyx mori : What do you want? A general way of attacking limits of ratios of functions in a wide class of functions (non-trivial smooth functions with taylor expansion equal to zero)? I mean, I can't even begin to imagine how many examples I could pull off out of my hat, and the exponential ones above are just the first one that came out. My point was to try to say that you shouldn't expect anything magical enough to satisfy a calculus student, even though the question is very legitimate on its own.
1d
answered When does L' Hopital's rule fail?
1d
answered Divisors of factorials
1d
answered Is the function $x \mapsto \max\{f(x),g(x)\}$ always an automorphism if $f$ and $g$ are?
1d
comment Is functional analysis a tool to prove uniform convergence?
Uniform convergence of what? Functional analysis is a tool for many things. Can you detail your question a bit more?
Aug
26
answered Group Theory $Z_2$ representations
Aug
26
comment What kind of space is this: $\Bbb{R}^n\times\Bbb{S}_{++}^n$?
What do you want to do with this space? If you don't want to do anything with it, then these questions don't mean much. Why did you consider this space in the first place?
Aug
26
comment Relationship of multiple particles under collision
Shouldn't this be on physics.stackexchange.com?
Aug
25
comment When a group acts on a metric space isometrically, transitively, and faithfully, is it a closed subspace of the isometry group?
So essentially you found a compact space $M$ with isometry group isomorphic to $G \times H$ where $G$ acts transitively and you took a non-closed subgroup of $H$, say $K$, so that $G \times K$ is not closed in $G \times H$. This example involves quite a lot of computations... I don't wanna go through them! OP should tell us if he did.
Aug
25
comment When a group acts on a metric space isometrically, transitively, and faithfully, is it a closed subspace of the isometry group?
@Martín : In $\mathbb R$ with the standard metric, the isometry group is isomorphic to $\mathbb R \times \{1,-1\}$ ($\pm 1$ if you flip the line or not). This is not compact.
Aug
25
comment How to create a new binary operation on a same set?
@AnjanDebnath : If $H$ has a group structure then you can transport it to $G$, that is what I am doing.
Aug
25
comment When a group acts on a metric space isometrically, transitively, and faithfully, is it a closed subspace of the isometry group?
What group $G$? I don't even know what action you are talking about ; there is a group action in the OP's statement.
Aug
25
comment How to create a new binary operation on a same set?
I use this bijection to take the group structure of $H$ and put it on $G$. If you pre-compose this bijection with a bijection from $G$ to itself, you can get different group structures on $G$, in particular by moving the identity element around. You can always start with a group $G$ and a bijection from $G$ to itself, and this will give you different (but isomorphic) group structures on $G$.
Aug
25
comment How to create a new binary operation on a same set?
@AnjanDebnath : You can always make sure the operation you get from $H$ on $G$ is a different one by ensuring that your bijection maps the identity of $H$ to an element different from the identity of $G$. If $G$ nor $H$ are groups, I have no operation from $H$ to pullback on $G$ (which is essentially what I did). The idea is to take a group $H$ which you have, and for which $|G| = |H|$ (i.e. there exists a bijection $\varphi : G \to H$).
Aug
25
answered How to create a new binary operation on a same set?