User abhinav kumar - MathOverflow

Answer by Abhinav Kumar for Morphsim from F_p[[X_1,...,X_d]].

2013-06-19T14:46:04Z

I think this is possible (over any base field $k$, not just $\mathbb{F}_p$). You can rephrase your question to ask: if an isomorphism $\phi: k[[x_1, \dots, x_d]]/I \to k[[y_1, \dots, y_d]]/J$ is specified, when can it be lifted to an isomorphism $k[[x_1, \dots, x_d]] \to k[y_1 , \dots, y_d]]$? Or even further, if you're given a surjection $$\pi: A = k[[x_1, \dots, x_d]] \to k[[y_1, \dots, y_d]]/J = B,$$ when does it come from an isomorphism $$\widetilde{\pi} : A \to C = k[[y_1, \dots, y_d]]?$$ Perhaps this follows from some "formal" properties, but in any case it's not a difficult exercise. You work with the map on tangent spaces $$ \overline{\pi} : {\frak{m}}_A/{\frak{m}}_A^2 \to {\frak{m}}_B/ {\frak{m}}_B^2 \cong {\frak{m}}_C/ (J \cap {\frak{m}}_C + {\frak{m}}_C^2) $$ which must be surjective. So you just have to extend it to an isomorphism of the $d$-dimensional vector spaces, by choosing an isomorphism of $K = \mathrm{ker}(\overline{\pi})$ with $(J \cap {\frak{m}}_C + {\frak{m}}_C^2)/ {\frak{m}}_C^2$, and then it will automatically induce an isomorphism $\widetilde{\pi}$.

Answer by Abhinav Kumar for Classification of these Binary Quadratic Forms

2013-06-12T14:14:51Z

For $x^2 + y^2$, the discriminant $b^2 - 4ac$ is $-4$, and the class number $h(-4)$ is $1$. So the condition is just that the discriminant matches up: $b^2 - 4ac = -4$, and the form is positive definite ($a, c > 0$). For $x^2 - y^2$, I think you just have to differentiate it from $2xy$, which also has discriminant $4$. So the condition is $b^2 - 4ac = 4$, and $a,c$ not both even. You can make the determinant of the transformation matrix $1$ in both cases, because the form is improperly equivalent to itself, for instance by $(x,y) \to (-x,y)$. For an elementary introduction see Chapter 3 of Niven, Zuckerman and Montgomery. For a more advanced treatment of quadratic forms, Cassels book "Rational quadratic forms" is a good reference. There's also Chapter 15 of Conway and Sloane's book "Sphere packings, lattices and groups".

Answer by Abhinav Kumar for Minimal representation of a polynomial as a linear combination of squares

2013-06-05T12:15:41Z

Ok, here's how to write a polynomial $f$ of degree $2n$ as a linear combination of three squares. We can assume $f$ is monic, since everything can be scaled at the end. Now, complete squares, starting from the top degree term. So we can write $f = g^2 + h$, where $g$ is monic of degree $n$, and $h$ has degree at most $n-1$. Then write $h = \frac{1}{4}(h+1)^2 - \frac{1}{4}(h-1)^2$ to finish it off. Three squares is optimal: a little bit of work shows that $x^4 + x + 1$ cannot be written as a linear combination of two squares (the same holds for a generic quartic polynomial). I guess this doesn't fully answer your second question of how to characterize linear combinations of two squares of degree $n$.

Answer by Abhinav Kumar for Tensor product of lattices

2013-06-03T20:17:27Z

You can consider the lattices embedded in Euclidean spaces $\mathbb{R^m}$ and $\mathbb{R^n}$ say. Then $U \otimes V$ is in $\mathbb{R}^m \otimes \mathbb{R}^n$ and the bilinear pairing (i.e. dot product) on this vector space is given by saying $\langle x \otimes y, x' \otimes y' \rangle = \langle x,y \rangle \cdot \langle y, y' \rangle$ for split tensors, and extending linearly to get it on the tensor product. If you don't want to consider embeddings into Euclidean spaces, just note that the lattice structure is equivalent to a quadratic or a bilinear form on the free abelian group. Then this formula defines the inner product for the tensor product. See, for instance, Section 1.10 of Martinet's book Perfect lattices in Euclidean spaces.

Answer by Abhinav Kumar for Question about the elementary divisors of a special matrix

2013-06-02T15:26:46Z

First consider the square case $n = k$. Let $A$ be the $n \times n$ matrix whose $(i,j)$ entry is $\phi(j)$ if $j | i$ and $0$ otherwise. Let $B$ be the matrix whose $(i,j)$ entry is $1$ if $i | j$ and $0$ otherwise. Now $M = AB$ follows from $$\sum_{k | \gcd(i,j)} \phi(k) = \gcd(i,j).$$ (More generally, you can replace $\gcd(i,j)$ by any natural number $m$ in the above identity.) Ok, now note that $A$ is lower triangular and $B$ is upper triangular. And in fact $B$ is in $GL_n(\mathbb{Z})$. So the elementary divisors of $AB$ are the same as $A$, which can be read off from the diagonal elements which are $\phi(1), \dots, \phi(n)$. (They're not the same as the the diagonal elements: you have to factor and rearrange so that $d_1 | d_2 | \dots | d_n$.)

For the more general situation, where you have a $k$ by $n$ matrix, assume $k \leq n$. Then define $A$ to be a $k \times k$ matrix and $B$ to be a $k \times n$ matrix, in similar fashion. The first $k \times k$ minor of $B$ is invertible over $\mathbb{Z}$, so you get the same elementary divisors as in the diagonal case, I believe.

Answer by Abhinav Kumar for Representation quaternions as matrices

2013-06-01T12:45:51Z

You could take the regular representation (left multiplication on $A$). So if $x^2 = a, y^2 = b$ then taking a basis ${1,x,y,xy}$ of $A$, $x$ would be represented by the matrix $$\left( \begin{array}{cccc} 0 & a & 0 & 0 \cr 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & a \cr 0 & 0 & 1 & 0 \end{array} \right), $$ etc.

Answer by Abhinav Kumar for zeros of a homogeneous polynomial

2013-05-30T22:08:01Z

Let $t$ be a root of $t^3 + \lambda t + 1 = 0$. Suppose $\lambda$ can be chosen so this polynomial is irreducible. Then the cubic decomposes into three lines, one of which is given by $x = ty + z/t$ (and the other two are the Galois conjugates). Now it's easy to see this line has no $\mathbb{F}_p$-rational points (else $t$ would satisfy an equation of degree less than $3$). Similarly the others don't have rational points either. So all you have to do is ensure that this poly is irreducible, and also that $4\lambda^3 + 27$ is nonzero, so you don't have a singular point over the base field.

Answer by Abhinav Kumar for Diophantine equation with primitive nth root of unity

2013-05-16T00:48:08Z

The expression $(- (\xi^k - 1)/(\xi - 1))^n$ is real: it equals the $n$'th power of $\sin(k \pi/n)/\sin(\pi/n)$ times $(-1)^{n+k-1}$, if I calculated correctly. So the thing you're taking absolute value of must be $\pm 1$. If it's $1$, you get the trivial solution. If it's $-1$, that tells you an $n$'th root of $\pm 2$ must be in the cyclotomic field $\mathbb{Q}(\xi_n).$ But the degree of the field obtained by adjoining a root of the polynomial $x^n \pm 2$ is $n$ (since the poly is irreducible by Eisenstein), and so this leads to $n \leq \phi(n)$, which is a contradiction for any $n > 1$.

Answer by Abhinav Kumar for orthogonality in a lattice

2013-02-27T14:36:00Z

For each $\lambda \neq 0$ in $\Lambda$, consider the subspace $H_\lambda$ of $W$ given by $\langle \lambda, x \rangle = 0$. Saying $W^\perp \cap \Lambda = 0$ means these are all proper subspaces of $W$. Now, since $\mathbb{R}$ is uncountable, it's not too hard to show that a countable union of proper subspaces can't be the whole vector space. So there's an element $w \in W$ which is outside the union, and therefore $w^\perp \cap \Lambda = 0$.

Answer by Abhinav Kumar for orthogonal base in unimodular lattice

2013-02-27T02:53:16Z

Given that $\Lambda$ is unimodular and indefinite, this can be done if and only if $\Lambda$ is odd (i.e. iff the diagonal entries $a_{i,i}$ are not all even). This follows from Milnor's classification. A couple of references where this is worked out are Serre's "A course in arithmetic" and Milnor and Husemoller's "Symmetric bilinear forms".

Answer by Abhinav Kumar for Distributing fire stations in a circular city

2013-02-24T20:53:19Z

This is the (finite) covering problem in the plane. According to Boroczky's book "Finite packing and covering", the answer is only known (provably) up to $k = 10$, due to the work of K. Bezdek and G. Fejes Toth. I don't know if the solutions are a famous sequence of geometric graphs.
_{(Image from MathWorld added by J.O'Rourke)}

Answer by Abhinav Kumar for Decomposition of primes in Galois closures of number fields

2013-02-13T12:19:13Z

With regard to the group-theoretic question, the answer is no: for example, there are plenty of faithful transitive group actions of different groups on the same set (i.e. only one orbit, but it doesn't determine the size of the group). For instance, take $S_n$ and $A_n$ acting on ${1,2,\dots, n}$, as soon as $n \geq 3$.

Answer by Abhinav Kumar for Reference for Complex Abelian Varieties

2013-02-12T05:09:24Z

Perhaps not in his "Abelian varieties" book, but certainly in his "Tata lectures on Theta" Mumford describes this setup. See Chapter II of book 1. Another reference is Birkenhake and Lange's book "Complex abelian varieties" (see Chapters 3 and 8, for instance). By the way, I don't think you generally construct just $g$ independent meromorphic functions; you construct a whole bunch more of them - enough to embed the abelian variety into projective space.

Answer by Abhinav Kumar for Efficient algorithm finding 'a' solution of system of linear inequalities

2013-02-11T23:13:09Z

You can do linear programming (for example, by giving it a random objective function); this will return a feasible solution - in fact, a vertex of the solution set. You can run such a linear program in maple (and possibly in sage). For something faster see the exact rational LP solver at http://www2.isye.gatech.edu/~wcook/qsopt/ex/index.html

Answer by Abhinav Kumar for Why the Abel-Jacoby map is algebraic morphism?

2013-02-11T21:05:12Z

Maybe there are easier ways to see it, but Chow's theorem/GAGA certainly gives you the result, since you have an analytic morphism of projective analytic varieties.

Answer by Abhinav Kumar for Solving an equation

2013-02-10T07:18:53Z

You can rewrite it as $(2t + 1)^2 = 8n^2 - 7$, so it's essentially equivalent to finding solution for the Pell equation $x^2 - 8y^2 = -7$. This has one solution $(1,1)$, and so it certainly has infinitely many. For example, the solution $(x,y) = (379, 134)$ gives you $n = 134, t = 189$. See, for instance, "An introduction to the theory of numbers" by Niven, Zuckerman and Montgomery, Section 7.8.

Answer by Abhinav Kumar for To what extent does trajectory determine gravity sources?

2013-02-09T02:25:54Z

Taking a cue from Joel's comment: it's easy to consider a trajectory going along the z-axis. A system of unit masses at $(1,0,0)$, $(-1,0,0)$, $(0,1,0)$ and $(0,-1,0)$ will give the same $(0,0,z)-t$ trajectory as for the system of unit masses at $(1,0,0)$, $(-1,0,0)$, $(\cos \theta, \sin \theta, 0)$ and $(-\cos \theta, -\sin \theta, 0)$.

Answer by Abhinav Kumar for linear independence of orbits via a set of transformations in char p

2013-02-06T04:50:52Z

Not in general: here is a counterexample. Take $p = 2$, and consider the matrices $T_1 = [1,0,0; 0,0,0; 0,0,0], T_2 = [0,0,0; 0,1,0; 0,1,0], T_3 = [0,0,0; 1,0,0; 0,1,0]$. (I'm using semi-colons to separate rows; a bit of LaTeX trouble formatting the matrix ...)

Then if $v$ is the column vector $(x,y,z)$, we get that the matrix $(T_1 v, T_2 v, T_3 v)$ is $[x,0,0; 0,y,x; 0,y,y]$. It has determinant $xy^2 - yx^2$. Note that if $x,y$ are in $\mathbb{F}_2$, then $x^2 = x$ and $y^2 = y$, so this determinant is $0$. However, the polynomial $xy^2 - yx^2$ is not identically $0$, so there are values in $\overline{\mathbb{F}_2}$for which the determinant will not vanish.

Update:

These transformations are not invertible, but you can embed this example in 4 dimensions. Namely, define $T_1(x,y,z,w) = (x,y,z,w)$, $T_2(x,y,z,w) = (x,x+y,z,w)$, $T_3(x,y,z,w) = (x,y,y+z,y+w)$ and $T_4(x,y,z,w) = (x,y,x+z, y + w)$. Then these are all clearly invertible, and the determinant of the matrix of linear forms is $x^2 y(y - x)$, which vanishes as before if $x$ and $y$ are in $\mathbb{F}_2$, but not over the algebraic closure.

Answer by Abhinav Kumar for orbits in tensor representations of GL(V)

2013-02-04T12:43:19Z

See, for instance, Chapter 5 of Goodman and Wallach's book "Symmetry, Representations and Invariants". Another good reference is Procesi's "Lie Groups: An approach through Invariants and Representations".

Answer by Abhinav Kumar for A curious sum for integers $\equiv 7\pmod 8$.

2013-02-01T14:44:44Z

There are probably lots of ways to see this, but here's one: let $S_1$ be the sum above, and let $$S_2 = \sum_{k=(n+1)/2}^{n-1} k \left( \frac{k}{n} \right).$$ Then $$S_1 + S_2 = S = \sum_{k=1}^{n-1} k \left( \frac{k}{n} \right).$$ Now you can rewrite $S$ (since $x \to 2x$ is a bijection mod $n$) as $$S = \sum_{k=1}^{(n-1)/2} 2k \left( \frac{2k}{n} \right) + \sum_{k=(n+1)/2}^{n-1} (2k-n) \left( \frac{2k}{n} \right) = 2S - n \sum_{k=(n+1)/2}^{n-1} \left( \frac{k}{n} \right), $$ where we used $(2/n) = 1$. Finally, we have $$S = 2S + n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right),$$ by changing $k$ to $n-k$ in the summation and using $(-1/n) = -1$. So $S = -n\sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right)$.

On the other hand, we can switch the index in $S_2$ from $k$ to $n-k$ as well, to get $$S_2 = \sum_{k=1}^{(n-1)/2} (n-k) \left( \frac{n-k}{n}\right) = -n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right) + \sum_{k=1}^{(n-1)/2} k \left( \frac{k}{n} \right) = S + S_1 = 2S_1 + S_2.$$ (We used $(-1/n) = -1$ again).

Therefore $S_1 = 0.$

How to show modularity of an elliptic curve?

2011-02-13T05:28:41Z

In the days before [W, TW, BCDT], how did people show that specific elliptic curves over $\mathbb{Q}$ were modular? For instance, I was reading through a paper of Buhler, Gross and Zagier from 1985 on the curve 5077a, and they say that modularity can be checked by a finite computation in the 422-dimensional space of cuspforms of weight 2 and level 5077 (and remark at the end that Serre and Mestre have checked it). A google search brought up the name "Faltings-Serre method": was this the technique of choice? Also, are there any good references for it?

Answer by Abhinav Kumar for A 'generalized Four Squares Theorem'?

2011-08-21T13:26:25Z

Yes, both are true. For example, see Theorem 1.6 of Chapter 11 of Cassels's "Rational Quadratic Forms", which says that if $Q$ is a positive definite integral quadratic form, then there is an integer $N$ depending on $Q$ such that if $a > N$ and $a$ is represented primitively by $f$ over all $\mathbb{Z}_p$ then $a$ is represented by $Q$. The local primitive representability is easy to show using the fact that $f$ is unimodular, and the classification of forms over $\mathbb{Z}_p$ by invariants.

For instance, if $p$ is odd and $p \nmid a$, then $f$ is equivalent over $\mathbb{Z}_p$ to $(a, a \det(f), 1, 1, \dots, 1)$, which obviously represents $a$ primitively. If $p | a$ you could use $((a-1), (a-1) \det(f), 1, 1, \dots, 1)$ which represents $a$ primitively. I won't do the analysis for $p = 2$, but see section 4 of chapter $8$ of Cassels.

Linear reductivity of $SL_n$ in char $0$: proof in Mukai's book

2011-08-10T04:38:18Z

I'm reading through Mukai's excellent book "Introduction to Invariants and Moduli", and am stuck on a proof in Chapter 4. He's proving that $G = SL_n$ over a field $k$ of characteristic $0$ is linearly reductive, i.e. for every epimorphism $V \rightarrow W$ of representations of $G$, the induced map on invariants $V^G \rightarrow W^G$ is also surjective. Let $\rho$ be a representation of $SL_n$ and let $\tilde{\rho}$ be the induced representation on the Lie algebra and the distribution algebra at the identity. Let $\Omega$ be the Casimir element/operator. Let $T$ be the torus of diagonal matrices in $SL_n$ and let $\frak{h}$ be its Lie algebra.

Mukai reduces the proof of linear reductivity to the following assertion: if $\mathrm{tr}( \tilde{\rho}(\Omega)) = 0$ we must also have $\mathrm{tr}(\tilde{\rho}(h)) = 0$ for all $h \in \frak{h}$. He then says: we will do this just for $SL_2$; the general case is similar. For $SL_2$ we have $\frak{h}$ is one-dimensional spanned by multiples of the root $h = \epsilon_1 - \epsilon_2$, and by explicit calculation $\mathrm{tr}( \tilde{\rho}(\Omega)) = \mathrm{tr}(\tilde{\rho}(h)^2)$. So the assertion is immediate. But I don't quite see how the "similar" proof for $SL_n$ works. It would be great if someone could explain this to me!

Answer by Abhinav Kumar for A good book of functional analysis

2011-08-09T01:51:17Z

John B. Conway's "A course in functional analysis" is also pretty decent.

Answer by Abhinav Kumar for Less discriminating discriminants

2011-08-07T02:14:20Z

Here's one way to think of this. Let $\lambda$ be a new variable and let $Q(z) = \lambda P'(z) + P''(z)$. Let $R(\lambda)$ be the resultant of $P$ and $Q$. Then we want $R$ to identically vanish (i.e. want to see if there's value of $z$ for which $P$ and $Q$ have a common zero, for every value of $\lambda$). So the coefficients of $R$ will describe an ideal cutting out the variety you want. This argument should work at least in characteristic $0$.

Answer by Abhinav Kumar for Lorentzian characterization of genus

2011-07-19T15:58:54Z

A good reference for this assertion is Cassels's "Rational Quadratic Forms", though you have to dig a bit. Let me see if I can outline the proof. First, I think Conway and Sloane assume $f$ and $g$ are classical integral (i.e. correspond to even lattices). In my copy of SPLAG, at the end of subsection 2.1 of that chapter, they say "so in this book we call $f$ an integral form if and only if its matrix coefficients are integers (i.e. if and only if it is classically integral ...)".

Now suppose $f$ and $g$ are in the same genus. Then so are $f\oplus H$ and $g \oplus H$. Next, we want to show they're in the same spinor genus. This follows from the Corollary of Lemma 3.6 of Chapter 11 of Cassels: "If we show $U_p \subset \theta(\Lambda_p)$ for all $p$, then the genus of $\Lambda$ consists of a single spinor genus". Here $\Lambda = f \oplus U$, where I'm identifying the form and the lattice by a bit of abuse of notation. Since $\theta(\Lambda_p) \supset \theta(H_p)$ (see a few sentences below the corollary), and $\theta(H_p) \supset U_p$ by Lemmas 3.7 and 3.8, we've proved that the genus consists of a single spinor genus.

Finally, since the forms are indefinite of dimension at least $3$, the spinor genus consists of a single class.

To go back is the easier direction (I think): if $f \oplus U$ is equivalent to $g \oplus U$, then they are equivalent over $\mathbb{Z}_p$ for every $p$. Then an analogue of Witt cancellation will do the job (see Chapter 8 of Cassels).

Answer by Abhinav Kumar for Hyperbolic Coxeter polytopes and Del-Pezzo surfaces

2011-07-19T14:41:04Z

Perhaps the nice article "Reflection groups in Algebraic Geometry" of Dolgachev describes this: link http://www.math.lsa.umich.edu/~idolga/reflections.pdf

Other places to look: papers of Vinberg and Nikulin, and Manin's book on "Cubic forms".

Answer by Abhinav Kumar for Most memorable titles

2011-05-05T19:58:58Z

The weird and wonderful chemistry of audioactive decay, by John Conway.

Answer by Abhinav Kumar for What is the Cayley projective plane?

2011-02-20T15:10:50Z

There's a very pretty construction using the exceptional Jordan algebra of dimension 27, described in the book "On Quaternions and Octonions" by Conway and Smith, as well as in Baez's article on the Octonions (see http://math.ucr.edu/home/baez/octonions/node12.html). To summarize, you take Hermitian 3 by 3 matrices over the Octonions, with Jordan product $A \circ B = (AB + BA)/2$. That gives you the exceptional Jordan algebra. If you further restrict to matrices which are of unit trace and idempotent, you get $\mathbb{OP}^2$. You say point $P$ lies on line $L$ if $P \circ L = 0$.

As some of the answers above have pointed out, the usual construction of higher dimensional projective spaces doesn't work because the Desargues theorem holds automatically in these (see Courant and Robbins, "What is Mathematics", pg. 171, for a nice illustration and quick proof), and would imply that $\mathbb{O}$ is associative, which it isn't. Note that the plane $\mathbb{OP}^2$ is non-Desarguian.

Answer by Abhinav Kumar for Proof synopsis collection

2011-02-15T05:55:30Z

Minkowski's lower bound for density of sphere packings in $\mathbb{R}^n$: take any sphere packing where you can't cram in any more spheres. Then doubling the size of the spheres must cover all space, which gives a lower bound of $\frac{1}{2^n}$.

Comment by Abhinav Kumar

2013-06-18T19:23:42Z

Can you make your question more mathematically precise?

Comment by Abhinav Kumar

2013-06-16T03:05:09Z

$GL_n$ is certainly a rational variety (over any field), since it's birational to $M_n$ which is affine $n^2$ space.

Comment by Abhinav Kumar

2013-06-08T13:07:43Z

Yes, volume is the volume of the fundamental domain (i.e. the lattice unit cell). Center density is the sphere packing density of the lattice divided by the volume of a unit ball. It's given by $r^n/V$ where $r = \min \Lambda /2$ is the radius of the sphere packing, and $V$ is the volume of the lattice. "Thickness" refers to the thickness of the lattice covering, i.e. the ratio $v R^n/V$, where $v$ is the volume of the unit ball, $R$ the covering radius, and $V$ is the lattice volume as above.

Comment by Abhinav Kumar

2013-06-07T01:51:26Z

In the situation above, I'm thinking of $AB$ acting on $\mathbb{Z}^n$ (column vectors), and we want to understand the cokernel of this matrix. By itself, $B$ has cokernel zero, so we just have to understand the cokernel of $A$.

Comment by Abhinav Kumar

2013-06-05T21:12:35Z

@Per Alexandersson - sorry, we posted comments almost simultaneously, so I didn't see your comment ...

Comment by Abhinav Kumar

2013-06-05T21:11:29Z

Compute a Grobner basis? (should be built-in in sage, if not then you can use the command from singular)

Comment by Abhinav Kumar

2013-06-05T16:33:32Z

As an example, $x^4 + 2x^3 + 3x^2 + 5x + 7 = (x^2 + x)^2 + 2x^2 + 5x + 7 = (x^2 + x + 1)^2 + 3x + 6$. Think of this as taking square roots near the place $x = \infty$ on the projective line.

Comment by Abhinav Kumar

2013-06-04T17:43:39Z

Are you looking for Hodge isometries between the $H^2$ or just between the transcendental lattices? (If the former, any Hodge isometry can be converted to an isomorphism of K3 after composing by a Weyl group element.)

Comment by Abhinav Kumar

2013-06-04T03:24:35Z

If I'm understanding Pourchet's result correctly, it seems to me that one should be able to represent any polynomial $f$ of even degree as a linear combination of six squares. We can assume $f$ is monic, then if you add a sufficiently large constant $c^2$ (with $c$ rational) to $f$, it will become non-negative everywhere: just take $c^2$ larger than the minimum value of $f$, noting $f$ goes to $\infty$ as the argument goes to $\pm \infty$. So by Pourchet, $f + c^2$ is a sum of at most five squares, and $f$ is a linear combination (with $\pm 1$ coeffs) of six squares. I doubt this is optimal!

Comment by Abhinav Kumar

2013-06-03T20:08:48Z

@David: Thanks - I'd forgotten that the inverse of $\zeta(s)$ has a nice L-series too! :-)

Comment by Abhinav Kumar

2013-05-29T21:24:47Z

The line breaks are all messed up, but I'm sure you can unentangle them.

Comment by Abhinav Kumar

2013-05-29T21:23:48Z

@Barry, you can get the recurrence by doing some linear algebra. Here's some gp code to do the m = 3 line (my indices are shifted by 1, since vectors and matrices in gp start with 1 rather than 0). Then you prove it by induction, of course :-) d = 80; A = matrix(d,d); for(i=1,d,A[i,1] = 1; A[i,2] = 1; A[1,i] = 1); for(i=2,d, for(j=3,d, A[i,j] = (A[i,j-2]*A[i-1,j] + A[i,j-1]*A[i-1,j-1])/A[i-1,j-2])); v = A[4,] m = matrix(40,9,i,j,v[i+j-1]); k = matkerint(m) p = vector(9,i,x^(i-1))*k factor(p[1])

Comment by Abhinav Kumar

2013-05-29T16:08:25Z

No worries! MO has its quirks.

Comment by Abhinav Kumar

2013-05-29T15:38:32Z

It looks correct now. I would suggest appending to your answer rather than editing it completely, so this comment thread makes sense ...

Comment by Abhinav Kumar

2013-05-29T15:14:15Z

There's a problem with your counterexample: namely that the span of A doesn't just consist of nilpotent matrices ($E_{1,2} + E_{2,1}$ squares to the identity, or to a projection matrix if the dimension is larger than $2$).