bio | website | dictionary.reference.com/… |
---|---|---|
location | Paris | |
age | 29 | |
visits | member for | 5 years, 3 months |
seen | 13 hours ago | |
stats | profile views | 969 |
In 2014, I obtained my PhD in math at CUNY Graduate Center. I enjoy mathematics.
In the song "Watching the Wheels," J. Lennon sings "There are no problems, only solutions." However, I prefer the simple Hegelian reversal of this statement: There are no solutions, only non-problems.
May 25 |
awarded | Nice Answer |
May 24 |
revised |
John Nash's Mathematical Legacy
added 3 characters in body |
May 24 |
revised |
John Nash's Mathematical Legacy
added 3 characters in body |
May 24 |
answered | John Nash's Mathematical Legacy |
May 8 |
comment |
When is the Hom-scheme connected?
oh, I thought of a trivial example: $Hom_k(spec(k), spec(k) \sqcup spec(k)) $. I think $A$ and $B$ should local artinian rings with residue field $k$ (or at least a finite extension of $k$). Then, I truly do not know of example having worked with such things for a few years. |
May 7 |
comment |
When is the Hom-scheme connected?
I like this question. When A and B are (truncated) infinitesimal neighborhoods of a smooth point, then $Hom$ will be connected (in fact, it will be iso to some $\mathbb{A}^M$). But, at non-singular points, explicit computations show that the space will often not be irreducible (yet remain connected). Off the top of my head at the moment, I cannot think of an example where $Hom$ is disconnected. The same question may be asked for non finite algebras, if you are willing to work with formal schemes. |
Jan 16 |
revised |
Why do we need localization by Leftschetz motive?
edited body |
Jan 16 |
answered | Why do we need localization by Leftschetz motive? |
Jan 16 |
comment |
Why do we need localization by Leftschetz motive?
Just to nit-pick, in any commutative ring $R$ and given any element $f$ of $R$, $R_f=$ the localization with respect to the stably multiplicative subset $\{1,f, f^2,\ldots\}$ is a ring. In particular $f^{-1}$ always exists in $R_f$ and is nonzero whenever $f$ is not nilpotent. |
Jan 3 |
comment |
Why is the CM closure of $\mathbb{Q}$ the “ultimate” coefficient field for motives?
This is a guess: it probably has something to do with periods. From section 3 of maths.ed.ac.uk/~aar/papers/kontzagi.pdf, if $f$ is a modular form of positive weight $k$ and $z_0\in\mathbb{H}$ is a CM point, then $\pi^kf(z_0)$ is a period. Among other things, there is a connection between nori motives and periods. |
Oct 20 |
comment |
Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$
Actually, this way is very easy also $\sum_{j=0}^{k-1} j^m = \frac{1}{m}(B_m(k) - B_m)$ which will immediately imply the result which you proved. The Bernoulli numbers were a red-herring. |
Oct 20 |
accepted | Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$ |
Oct 20 |
comment |
Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$
ah, yes, now the other hint makes sense too. Thanks for your help. Your answer implies that $(B_m(k)-B_m)/mn^m$ goes to zero as $m$ approaches infinity provided $n>1$. I thought that maybe there is a way to prove this second fact directly. |
Oct 20 |
comment |
Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$
Thanks for your hint. Yes, it is true, but the events $E_i$ given by $\{(l_j) \mid l_i = n\}$ are not mutually exclusive. I should say that neither probability theory nor analysis are my specialty so perhaps I missed the hint. |
Oct 20 |
asked | Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$ |
Sep 30 |
awarded | Explainer |
Aug 14 |
answered | Vanishing of Motivic Cohomology |
Aug 14 |
revised |
Vanishing of Motivic Cohomology
fixed a couple of typos |
Aug 14 |
suggested | approved edit on Vanishing of Motivic Cohomology |
Jan 1 |
awarded | Yearling |