User dipramit majumdar - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:31:09Z http://mathoverflow.net/feeds/user/4291 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16651/weil-group-weil-deligne-group-scheme-and-conjectural-langlands-group Weil group, Weil-Deligne group scheme and conjectural Langlands group Dipramit Majumdar 2010-02-28T03:42:00Z 2011-08-03T02:57:33Z <p>I was reading a series of article from the Corvallis volume. There are couple of questions which came to my mind:</p> <ol> <li><p>Why do we need to consider representation of Weil-Deligne group? That is what is an example of irreducible admissible representation of $Gl(n,F)$ which does not correspond to a representation of $W_F$ of dimension $n$ ? An example for $n=2$ will be of great help.</p></li> <li><p>In the setting of global Langlands conjecture, why extension of $W_F$ by $G_a$ or products of $W'_{F_v}$ does not work? </p></li> </ol> <p>Thank you.</p> http://mathoverflow.net/questions/20339/reference-for-unitary-group-attached-to-e-k Reference for Unitary Group attached to $E/k$ Dipramit Majumdar 2010-04-04T22:55:13Z 2010-04-05T07:05:15Z <p>Unitary groups are very important objects in the setting of Langland's Conjecture because of the existence of Shimura Variety ( which I don't know) and also because people know how to attach a galois representation to a automorphic representation of unitary groups in almost all cases (By the work of Taylor, Harris and many others)(Which also I don't know).</p> <p>I was trying to learn about unitary groups attached to $(k,E,D,*)$ where say $k$ is a totally real field,$E$ CM field of deg 2 over $k$, $D$ is a central simple algebra of rank $n^2$ over E, and $*$ $k$ algebra anti-involution of 2nd kind on $D$ (i.e. coinciding with the action of non-trivial element of $Gal(E/k)$ on $E$).</p> <p>But I could not really find a reference for this. Essentially some authors define it a inner form of a particular quasi-split unitary group, and some authors define it as functor of points. Also it is commented that there is some sort of Global-Local patching going on. Can any one give me a reference where unitary groups is covered in some what details rather than a overview in 2 pages?</p> http://mathoverflow.net/questions/20112/interesting-results-in-algebraic-geometry-accessible-to-3rd-year-undergraduates/20120#20120 Answer by Dipramit Majumdar for Interesting results in algebraic geometry accessible to 3rd year undergraduates Dipramit Majumdar 2010-04-02T00:29:47Z 2010-04-02T00:29:47Z <p>Bezout's theorem can be very entertaining. I saw a elementary proof (essentially using linear algebra and commutative algebra) of this as an undergrad.(Unfortunately I don't recall the proof or the reference, but here is a <a href="http://mcc1.mccfl.edu/fl_maa/proceedings/2001/fitchett.pdf" rel="nofollow">reference</a>) And the statement was quite astonishing to me at that point ( like 2 circles intersect in 4 points ...).Also it introduces the idea of intersection multiplicity. </p> http://mathoverflow.net/questions/19899/terminology-occuring-in-automorphic-representation-and-relationship-between-them Terminology occuring in automorphic representation and relationship between them Dipramit Majumdar 2010-03-31T00:05:04Z 2010-03-31T17:03:10Z <p>When one tries to read about automorphic representation few terms come up more than others namely,</p> <p>1.Cuspidal</p> <p>2.Square Integrable</p> <p>3.Absolutely Cuspidal</p> <p>4.Super Cuspidal</p> <p>My understanding about them is Cuspidal and Square Integrable Representation are the same. Older authors used the term Square Integrable, where as now days people use Cuspidal.</p> <p>Similarly Absolutely Cuspidal and Super Cuspidal Representation are same. Older authors used the term Absolutely Cuspidal, but now days people use the term Super Cuspidal to mean the same thing.</p> <p>What is the reason and history behind this change of terminology? Or am I completely wrong and each of them refer to different objects?</p> <p>Also what is the relationship between Cuspidal and Super Cuspidal Representation?</p> http://mathoverflow.net/questions/19684/relation-between-hecke-operator-and-hecke-algebra Relation between Hecke Operator and Hecke Algebra Dipramit Majumdar 2010-03-29T03:34:21Z 2010-03-30T02:12:16Z <p>In the study of number theory (and in other branches of mathematics) presence of Hecke Algebra and Hecke Operator is very prominent.</p> <p>One of the many ways to define the Hecke Operator $T(p)$ is in terms of double coset operator corresponding to the matrix $\begin{bmatrix} 1 &amp; 0 \ 0 &amp; p \end{bmatrix}$ . </p> <p>On the other hand Hecke Algebra $\mathcal{H}(G,K)$ associated to a group $G$ of td-type ( topological group, such that every neighborhood of unity contains a compact open subgroup), where $K$ is a compact open subgroup of $G$ is defined as the space of locally constant compactly supported $K$ bi-invariant functions on $G$. Convolution product makes it an associative algebra. </p> <p>I was told that the hecke algebra $\mathcal{H}(Gl(2,\mathbb{Q}_p) , Gl(2,\mathbb(Z)_p))$ corresponds to the classical algebra of hecke operators attached to $p$ via Satake Isomorphism Theorem. Using Satake Isomorphism theorem I can show $\mathcal{H}(Gl(2,\mathbb{Q}_p) ,Gl(2,\mathbb(Z)_p))$ is commutative and finitely generated over $\mathbb{C}$. </p> <p>So my question is how one uses Satake Isomorphism Theorem (or otherwise) to see this? And secondly in general what is the relation between hecke operators and hecke algebra?</p> http://mathoverflow.net/questions/19586/definition-of-l-function-attached-to-automorphic-representation/19664#19664 Answer by Dipramit Majumdar for Definition of L-function attached to automorphic representation Dipramit Majumdar 2010-03-28T20:44:48Z 2010-03-28T20:44:48Z <p>As far as I understand, attaching an $L$-function to an automorphic representation attached to a general reductive group $G$ is conjectural and still open.</p> <p>The way one attaches $L$-function depends on a representation $r$ of ${^L}G$ and partitioning the set of irreducible admissible representation of $G(k_v)$ in to $L$-packets (which is conjectural in general and known in very few cases). Assuming one can define local $L$-packets, Borel and Tate's article in Corvalis explains how to attach $L$-function to it. But still this $L$-function depends on the chosen representation $r$.</p> <p>If $\pi$ is an irreducible admissible representation of $G_A$ then $\pi= \otimes_v \pi_v$, where $\pi_v$ is an irreducible admissible representation of $G(k_v)$. So assuming we can partition the set of irreducible admissible representation of $G(k_v)$ in to $L$-packets, $\pi_v$ belongs to $L$-packet $\Pi_{\phi_v}$ corresponding to some admissible homomorphism $\phi_v$ of Weil-Deligne group to ${^L}G/k_v$ . The repesenation $r$ defines a representation $r_v$ of ${^L}G/k_v$.</p> <p>Then the $L$-function attached to $\pi$ and $r$ is defined as: $L(s,\pi,r) = \prod_v L(s,\pi_v,r_v)$, $L(s,\pi_v,r_v)=L(s, r_v \circ \phi_v)$</p> <p>Now $r_v \circ \phi_v$ is a represenatation Wiel-Deilgne group, so by Tate's article in Corvalis, we know local $L$-factor.</p> http://mathoverflow.net/questions/19899/terminology-occuring-in-automorphic-representation-and-relationship-between-them/19902#19902 Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-31T22:03:02Z 2010-03-31T22:03:02Z Thanks I got a better idea now.Much appreciated. http://mathoverflow.net/questions/19899/terminology-occuring-in-automorphic-representation-and-relationship-between-them/19902#19902 Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-31T02:00:20Z 2010-03-31T02:00:20Z I don't see how Proposition 5.1.1 and Thm 4.4.6 imply super-cuspidal representation are square integrable. The Proposition 5.1.1 is just restatement of the definition in terms of Jacquet's fuctor. http://mathoverflow.net/questions/19899/terminology-occuring-in-automorphic-representation-and-relationship-between-them/19902#19902 Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-31T01:26:14Z 2010-03-31T01:26:14Z I am quite certain that by Absolutely Cuspidal and Super Cuspidal are the same thing as far as automorphic representation are concerned. It is also used to be called Parabolic.(Reference: Definition 1.3 Cartier Notes in Corvalis titled Representation of p-adic Groups) Also I would be really surprised if Cuspidal and Super-Cuspidal representation are same. Super cuspidal representation are not usually square integrable, where as cuspidal representations are. http://mathoverflow.net/questions/19768/upper-half-plane-quotient-by-a-discrete-group Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-29T21:27:31Z 2010-03-29T21:27:31Z elliptic points under group action means usually means stabilizer of the point is a nontrivial subgroup (eg, \Gamma = \Gamma(1), then $i$ and $\pho=cube root of unity$ are elliptic points and parabolic fixed point means points stable under the action of parabolic elts(means absolute value of trace is 2, but element is not I or -I , eg [1 1; 0 1]), I usually think of them as cusps http://mathoverflow.net/questions/19684/relation-between-hecke-operator-and-hecke-algebra/19757#19757 Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-29T20:44:11Z 2010-03-29T20:44:11Z Your explanation of the relationship between Hecke Algebra and Hecke Operator is great. In fact it shows the relationship is much more basic than the Satake Isomorphism. My understanding of Statake Isomorphism was H(G//K) is the symmetric polynomials in 2 variables, rather than C[T,S,S^{-1}], as Weyl group is S_2. Probably I misunderstood,I will go back to Cartier article and unravell the definitions and isomorphisms .Thanks for this valuable comment and the answer. http://mathoverflow.net/questions/19684/relation-between-hecke-operator-and-hecke-algebra/19757#19757 Comment by Dipramit Majumdar Dipramit Majumdar 2010-03-29T19:33:27Z 2010-03-29T19:33:27Z This explanation is very useful.The place where I read about how to think of hecke operators adelicly was the book Automorphic forms on Adele groups by Gelbert. I recently came up with a version of Satake Isomorphism Theorem which says $\mathcal(H)(Gl(2,\mathbb(Q)_p),Gl(2,\mathbb(Z)_p)) \cong \mathbb(C)[T,S,S^-1]$, I was told that $T$ corresponds to the hecke operator $T(p)$ when thought as an operator on automorphic form.I am not familiar with this form of Satake isomorphism theorem, so was hoping someone can explain this.But your answer clearly shows there is a much elementary connection. http://mathoverflow.net/questions/16651/weil-group-weil-deligne-group-scheme-and-conjectural-langlands-group/16659#16659 Comment by Dipramit Majumdar Dipramit Majumdar 2010-02-28T06:54:59Z 2010-02-28T06:54:59Z Thanks a lot Matt. http://mathoverflow.net/questions/16651/weil-group-weil-deligne-group-scheme-and-conjectural-langlands-group/16656#16656 Comment by Dipramit Majumdar Dipramit Majumdar 2010-02-28T05:07:38Z 2010-02-28T05:07:38Z I see, since it corresponds to a nontrivial nilpotent operator, it can not come from a representation of Weil group. Or is there some other argument? Thanks http://mathoverflow.net/questions/16651/weil-group-weil-deligne-group-scheme-and-conjectural-langlands-group Comment by Dipramit Majumdar Dipramit Majumdar 2010-02-28T03:55:08Z 2010-02-28T03:55:08Z Corrected the spelling. Thanks