Dror Speiser
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Registered User
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Masters student at Tel Aviv University
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Apr 17 |
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Reconciling Lusztig’s results with the Langlands philosophy ... explaining why Digne-Michel (well, really Lusztig in his work) turn to the problem of describing $\mathcal{E}(C_\hat{G}(s),1)$. Essentially, your question would be answered if you could describe these Lusztig series with a part of an L-parameter, whether into $\hat{G}(\mathbb{C})$ or $\hat{G}(\mathbb{F}_q)$. |
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Apr 17 |
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Reconciling Lusztig’s results with the Langlands philosophy @Will: first of all, not exactly, as the characters I mentioned are the Deligne-Lusztig characters, and not the $R_{ss}$. Second, if you mention 8.4.6, then I'm sure you're also aware of 8.4.7, that also give the regular characters as a sum over Deligne-Lusztig characters. Together, these give all irreducible characters for a few low-dimensional groups, which should be the starting point of an answer to your question. Now, Lusztig's result is that you can parametrize irreducible characters by $\{\chi_{s,\lambda}\ |\ s\in \hat{G}\ \text{semisimple},\ \lambda \in \mathcal{E}(C_\hat{G}(s),1)\}$.. |
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Apr 16 |
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Reconciling Lusztig’s results with the Langlands philosophy ... This is remark 5.21.5 in the original Deligne-Lusztig paper, and also appears in Carter's Finite Groups of Lie Type in the chapter on geometric conjugacy. |
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Apr 16 |
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Reconciling Lusztig’s results with the Langlands philosophy @Will: I think you are missing something, though I'm not sure it gives an answer to the big Langlands-analogy question: namely $\hat{F}$-stable semisimple conjugacy classes of $\hat{G}(\mathbb{F}_q)$ are in bijection with geometric conjugacy classes of $(T,\theta)$ on $G$. These in turn give virtual representation - the Deligne-Lusztig characters. For the most part, these are in fact irreducible characters, and to understand their decomposition it is enough to understand the Lusztig series. This is the part I am less familiar with, so can't say much more. |
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Apr 12 |
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Status of (Global) Langlands Conjecture for $GL_2$ over $\mathbb{Q}$ @Joel: Yeah, I think so, isn't this a part of the Langlands-Tunnel theorem? |
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Apr 11 |
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Status of (Global) Langlands Conjecture for $GL_2$ over $\mathbb{Q}$ I hope I don't sound too ignorant, but doesn't this answer ignore the construction from (2) to (1c) for even solvable galois groups? |
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Apr 8 |
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Approximate number of primes below a given integer? I'm guessing the words "polynomial time" above are meant in the logarithm of $n$. To answer that, it would be interesting to know how well the Lagarias-Odlyzko algorithm approximates $\pi(x)$ if one is only willing to put it in $log^k(x)$ time. |
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Apr 5 |
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When does a modular form satisfy a differential equation with rational coefficients? expanded on idea |
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Mar 31 |
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When does a modular form satisfy a differential equation with rational coefficients? @robot: hey, thanks for the link. I've seen this paper before, and unfortunately (for me) it suffers from the same problem every paper I've seen suffers from: it begins with any given modular function $t$, instead of constructing an interesting one. |
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Mar 27 |
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When does a modular form satisfy a differential equation with rational coefficients? @Francois: thanks for the reference. My french is a bit a rusty (non-existent), but I think the remark says what I wrote above: that the coefficients are in general algebraic. |
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Mar 27 |
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When does a modular form satisfy a differential equation with rational coefficients? I also asked this in math.stackexchange: math.stackexchange.com/questions/338453/… |
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Mar 27 |
asked | When does a modular form satisfy a differential equation with rational coefficients? |
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Mar 5 |
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Which level structures on elliptic curves are twist-invariant? I guess if you want to cover 4th and 6th roots then you need the group to contain elements corresponding to $\mu_4$ and $\mu_6$, i.e. conjugate to $\left(\begin{array}{cc}0 & -1 \\ 1 & 0 \end{array}\right)$ and $\left(\begin{array}{cc}-1 & -1 \\ 1 & 0 \end{array}\right)$ |
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Mar 3 |
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Do isogenies with rational kernels tend to be surjective? The j-invariant of $E$ is: $$\frac{-(d^4 + 12d^3 + 14d^2 - 12d + 1)^3}{d^5(d^2 + 11d - 1)}$$ The j-invariant of $E'$ is: $$\frac{-(d^4 - 228d^3 + 494d^2 + 228d + 1)^3}{d(d^2 + 11d - 1)^5}$$ So, in a sense, $j(E)$ is just arithmetically simpler. For example, computation shows that the modular degree (using $X_0$) of $E$ is usually the smaller one of the two, so its Heegner points should have smaller height, making $\eta$ surjective. |
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Feb 25 |
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Is strong multiplicity one (obviously) stronger than multiplicity one? @Tom: I thought "strong multiplicity one" means that not only are the two representations isomorphic, but also equal (now clearly implying your "multiplicity one"). At least this is Theorem 3.3.6 in Bump's Automorphic Forms and Representations (for the case $G=GL_n$). In any case (like $GL_n$), if you have uniqueness of Whittaker models, then automorphic functions are determined by their Whittaker coefficients, so if two cuspidal representations are isomorphic - their functions will have the same Whittaker coefficients up to a constant, and hence the representations are in fact equal. I think. |
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Feb 18 |
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Find polynomial in finite field So, probably $O(k^w\text{poly}(\log (q) ))$. |
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Feb 4 |
awarded | ● Popular Question |
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Jan 18 |
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Square Roots of Unity modulo N^2 Do you really need Hensel's lemma? |
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Nov 24 |
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Most “natural” proof of the existence of Hilbert class fields Hey Franz, sorry for replying almost who years later. Saw this again while going through favourites. My point was that the first sentence of the second paragraph is false, by giving a counter example. In your reply you used an ideal $\mathfrak{a}=(1)$ that doesn't generate the class group of $K$, which you required in the last sentence of the first paragraph. |
