Aakumadula
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Registered User
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interested in representations of groups, algebraic groups, arithmetic groups, rigidity, and related automorphic forms. Recently interested in braid groups and configuration spaces.
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4h |
awarded | ● Enlightened |
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4h |
awarded | ● Nice Answer |
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May 18 |
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How to detect if a subgroup lands inside an orthogonal group? There are criteria to tell if the form that the "self-dual" group $H$preserves is orthogonal or symplectic, in terms of the action by a particular element of the centre of $H$. All this is classical, and I am sure is in standard textbooks. |
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May 18 |
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How to detect if a subgroup lands inside an orthogonal group? Yes, of course that is the case. The natural representation of the orthogonal group is self dual,hence so is its restriction to $H$. But not conversely: the symplectic group also has self dual natural representation. |
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May 18 |
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Are residually finite, perfect groups residually alternating? in the above comment, please change "have unipotent groups" to "have nilpotent groups as normal subgroups" |
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May 18 |
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Are residually finite, perfect groups residually alternating? @victor: yes this needs a proof. A finite quotient of $SL_n({\mathbb Z})$ , by CSP, is a quotient of $SL_n({\mathbb Z}/n{\mathbb Z})$. By strong approximation, the latter is the product of $SL_n({\mathbb Z}/p^e{\mathbb Z})$ for varying primes $p$ and integers $e$. The latter groups have unipotent groups and the unique simple quotient $SL_n({\mathbb Z}/p{\mathbb Z})$ (modulo centre, as you rightly say). |
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May 17 |
accepted | Are residually finite, perfect groups residually alternating? |
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May 17 |
answered | Are residually finite, perfect groups residually alternating? |
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May 16 |
accepted | Cyclotomic fields |
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May 16 |
answered | Cyclotomic fields |
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May 4 |
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Weyl law for arithmetic Fuchsian groups known? The only arithmetic, nonuniform cofinite Fuchsian groups are (after a conjugation in $SL_2({\mathbb R})$) commensurable with $SL_2({\mathbb Z})$. So the issue is congruence subgroups vs. non-congruence subgroups of $SL_2({\mathbb Z})$. |
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May 2 |
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j-invariant duplication, triplication and quintuplication formulae… how? added 128 characters in body; edited body; added 3 characters in body |
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May 2 |
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j-invariant duplication, triplication and quintuplication formulae… how? added 2 characters in body |
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May 2 |
answered | j-invariant duplication, triplication and quintuplication formulae… how? |
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May 1 |
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Generators of class groups Dear Guillaume, not at all! Yes, indeed, any class of $G$ can be represented by a prime ideal. This is analogous to Dirichlet's theorem on infinitude of primes in arithmetic progressions; indeed, that any element of $({\mathbb Z}/m{\mathbb Z})^*$ (* denotes the group of units) is represented by primes may be interpreted as saying that elements of certain generalized ideal classes are represented by primes. |
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May 1 |
accepted | Generators of class groups |
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May 1 |
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Generators of class groups Yes, indeed. the equality is attained when $G$ is abelian of exponent two. |
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Apr 30 |
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Generators of class groups added 13 characters in body |
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Apr 30 |
revised |
Generators of class groups added 36 characters in body |
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Apr 30 |
answered | Generators of class groups |
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Apr 29 |
awarded | ● Yearling |
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Apr 27 |
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Inequivalence of group representations preserved under tensor product? I mean the following; suppose $F$ is a finite group for which Richard Stanley's example works. Suppose $G$ is a Lie group whose connected component group $G/G^0$ is $F$. Then for $G$ also, there exist two reps $r_1,r_2$ as above. |
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Apr 26 |
accepted | Normal subgroup of the identity component of a linear Lie group is normal in the whole group? |
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Apr 26 |
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Inequivalence of group representations preserved under tensor product? I would say the difference comes from the connectedness or otherwise of the group. If $G$ is not conected, then the group of connected components may well be of the type given by Richard Stanley. |
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Apr 26 |
answered | Normal subgroup of the identity component of a linear Lie group is normal in the whole group? |
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Apr 25 |
revised |
Inequivalence of group representations preserved under tensor product? added the assumption that $r_1,r_2$ are irreducible. |
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Apr 25 |
accepted | Inequivalence of group representations preserved under tensor product? |
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Apr 25 |
answered | Inequivalence of group representations preserved under tensor product? |
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Apr 13 |
answered | Cool problems to impress students with group theory |
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Apr 11 |
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Calculation of $H^{2}(S^{2k-1}/G,\mathbb{Z})$ Under these conditions, the cohomology of the quotient (with real coefficients) is the space of $G$ invariants in the cohomology of the total space $S^{2k-1}$ and hence $H^2$ of the quotient , with real coeffts., vanishes. |
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Apr 11 |
awarded | ● Fanatic |
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Apr 7 |
revised |
Generators for a certain congruence subgroup of SL(n,Z) added 267 characters in body |
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Apr 6 |
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Generators for a certain congruence subgroup of SL(n,Z) @jean: your argument only proves the local statement that in $SL_n$ of the $l$-adic integers (suppose $l$ is a prime), the $e_{ij}^l$ generate the appropriate congruence closure. It does not prove that at the global level (i.e. the integral points of $SL_n$, ), the $e_{ij}^l$ generate the appropriate congruence subgroup. You do need the theorem of Tits (and that is a non-trivial result). |
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Apr 6 |
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Classification for a special simple group I am a bit puzzled. Any integer $m$ divides $q-1$ for a suitable $q=p^m$. So, the class of simple groups $G$ such that $Card(G)$ divides $q(q^2-1)/2$ is the class of all simple groups. You do not assume that $p$ divides the order of $G$, so can this question be phrased purely in terms of a simple group $G$? |
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Apr 6 |
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Generators for a certain congruence subgroup of SL(n,Z) I do not understand Cooper's remark. The theorem of Tits quoted above shows that $e_{ij}^l$ generate a congruence subgroup, which can only be $\Gamma _n(l^2)$. |
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Apr 4 |
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Is there non-simple-connected projective variety(over C) with trivial etale fundamental group? You have mentioned this result that every f.p.group is the fundamental group of a complex manifold. But is it true that it is the fundamental group of some algebraic variety? For example, if we take smooth projective varieties, their fundamental groups are restricted (for example, the abelianisation of $\pi _1$ must have even rank as an abelian group). |
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Apr 4 |
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Is there non-simple-connected projective variety(over C) with trivial etale fundamental group? I am confused about the question. a (smooth) Variety is different from a complex manifold. Are these examples of $G$ also the topological fundamental groups of varieties or only of complex manifolds? |
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Apr 4 |
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Generators for a certain congruence subgroup of SL(n,Z) @Edward : Bass-Milnor-Serre and its "antecedents" is not quite right. There s this paper of Tits (which uses Bass-Milnor-Serre) where he proves that the group generated by $e_{ij}^l$ is a finite index subgroup for $n\geq 3$. |
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Apr 3 |
answered | Generators for a certain congruence subgroup of SL(n,Z) |
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Mar 20 |
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Spectral synthesis for central functions on locally compact groups was your advisor laurent clozel? |
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Mar 20 |
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Principal maximal ideals in Z[x]/(F) I sort of agree, but it is not clear that for the order $A={\mathbb Z[x]/(F)$, the intersection of the principal maximal ideal with $A$ is actually principal even if it is so for the smaller ring ${\mathbb Z}+MO_K$. . |
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Mar 19 |
accepted | Galois groups of CM fields |
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Mar 19 |
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Galois groups of CM fields yes, thanks. I do mean $Gal(L_0)$.Corrected now. |
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Mar 19 |
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Galois groups of CM fields added 2 characters in body |
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Mar 19 |
answered | Galois groups of CM fields |
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Mar 19 |
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Galois groups of CM fields The Galois group is of the Galois closure of $K$. |
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Mar 19 |
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Galois groups of CM fields When you take $K_0={\mathbb Q}({\sqrt 2}$ and $K=K_0[X]/(X^2+{\sqrt 2})$, it is clear that $[K:{\mathbb Q}]=4$ but it is not Galois. The Galois group is not abelian but is of order $8$. |
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Mar 19 |
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Principal maximal ideals in Z[x]/(F) @ayanta thanks for this. I think you may want to put this up as an answer since it answers the question completely (in the negative). |
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Mar 19 |
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Intersection in the complexification of a lattice added 7 characters in body |
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Mar 18 |
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Principal maximal ideals in Z[x]/(F) @ayanta, I first thought you were right, but I am now confused. If $P\subset O_K$ is a maximal ideal whose residue char does not divide $[O_K:A]$, how can it lie in $A$? If $P$ lies in $A$, it means that $[O_K:P]$ is in fact a multiple of $[O_K:A]$. @martin brandenburg: $Z[x]/(F)$ is definitely an order (since its ${\mathbb Q}$ span is the number field and this is a subring of $O_K$. |
