Wednesday, February 24, 2016

Sound and color

Just back home from The Stone where I could hear two very interesting sets with pianist Russ Lossing and drummer Gerry Hemingway, first in duet, and then in quartet with Loren Stillman on alto saxophone and Samuel Blaser on trombone.

I was absolutely excited at the prospect of returning to this avant-garde jazz hall (it has been my 3rd concert there, the first one was in 2010, with Sylvie Courvoisier, Thomas Morgan and Ben Perowski, and the second, last year, with Wadada Leo Smith and Vijay Iyer) to listen to Gerry Hemingway, and the cold rain falling on New York City did not diminish my enthusiasm. (Although I had to take care on the streets, for one could almost see nothing...) I feared I would arrive late, but Gerry Hemingway was still installing his tools, various sticks, small cymbals, woodblocks, as well as a cello bow...

I admit, it took me some time to appreciate the music. Of course, it was free jazz (so what?) and I couldn't really follow the stream of music. Both musicians were acting delicately and skillfully (no discussion) at creating sound, as a painter would spread brush strokes on a canvas—and actually, Hemingway was playing a lot of brushes, those drum sticks made of many (wire or plastic) strings that have a delicate and not very resonating sound... Color after color, something was emerging, sound was being shaped.

There is an eternal discussion about the nature of music (is it rhythm? melody? harmony?) and consequently about the role of each instrument in the shaping of the music. A related question is the way a given instrument should be used to produce sound.

None of the obvious answers was to be heard tonight. Russ Lossing sometimes stroke the strings of the grand piano with mallets, something almost classical in avant-garde piano music. I should have been prepared by the concert of Tony Malaby's Tubacello, that I attended with François Loeser in Sons d'hiver a few weeks ago, where John Hollenbeck simultaneously played drums and prepared piano, but the playing of  Gerry Hemingway brought me much surprise. He could blow on the heads of the drums, hit them with a woodblock or strange plastic mallets; he could make the cymbals vibrate by pressing the cell bow on it; he could also take the top hi-hat cymbal on the left hand, and then either hit it with a stick, or press it on the snare drum, thereby producing a mixture of snare/cymbal sound; during a long drum roll, he could also vary the pitch of the sound by pressing the drum head with his right foot—can you imagine the scene?

It is while discussing with him in between the two sets that I gradually understood (some of) his musical conception. How everything is about sound and color. That's why he uses an immense palette of tools, to produce the sounds he feels would best fit the music. He also discussed extended technique, by which he means not the kind of drumistic virtuosity that could allow you (unfortunately, not me...) to play the 26 drum rudiments at 300bpm, but by extending the range of sounds he can consistently produce with his “basic Buddy Rich type instrument”—Google a picture of Terry Bozzio's drumkit if you don't see what I mean. He described himself as a colorist, who thinks of his instrument in terms of pitches; he also said how rhythm also exists in negative, when it is not played explicitly. A striking remark because it exactly depicted how I understand the playing of one of my favorite jazz drummers, Paul Motian, but whom I couldn't appreciate until I became able of hearing what he did not play.

The second set  did not sound as abstract as the first one. Probably the two blowing instruments helped giving the sound more flesh and more texture. Samuel Blaser, on the trombone, was absolutely exceptional—go listen at once for his Spring Rain album, an alliance of Jimmy Giuffre and contemporary jazz—and Loren Stillman sang very beautiful melodic lines on the alto sax. The four of them could also play in all combinations, and with extremly interesting dynamics, going effortlessly from one to another. And when a wonderful moment of thunder ended abruptly with the first notes of Paul Motian's Etude, music turned into pure emotion.

Tuesday, February 9, 2016

Happy New Year!

As was apparently first noticed by Noam Elkies, 2016 is the cardinality of the general linear group over the field with 7 elements, G=GL(2,F7)G=\mathop{\rm GL}(2,\mathbf F_7). I was mentoring an agrégation lesson on finite fields this afternoon, and I could not resist having the student check this. Then came the natural question of describing the Sylow subgroups of this finite group. This is what I describe here.

First of all, let's recall the computation of the cardinality of GG. The first column of a matrix in GG must be non-zero, hence there are 7217^2-1 possibilities; for the second column, it only needs to be non-collinear to the first one, and each choice of the first column forbids 77 second columns, hence 7277^2-7 possibilities. In the end, one has Card(G)=(721)(727)=4842=2016\mathop{\rm Card}(G)=(7^2-1)(7^2-7)=48\cdot 42=2016. The same argument shows that the cardinality of the group GL(n,Fq)\mathop{\rm GL}(n,\mathbf F_q) is equal to (qn1)(qnq)(qnqn1)=qn(n1)/2(q1)(q21)(qn1)(q^n-1)(q^n-q)\cdots (q^n-q^{n-1})=q^{n(n-1)/2}(q-1)(q^2-1)\cdots (q^n-1).

Let's go back to our example. The factorization of this cardinal comes easily: 2016=(721)(727)=(71)(7+1)7(71)=6876=253272016=(7^2-1)(7^2-7)=(7-1)(7+1)7(7-1)=6\cdot 8\cdot 7\cdot 6= 2^5\cdot 3^2\cdot 7. Consequently, there are three Sylow subgroups to find, for the prime numbers 22, 33 and 77.

The cas p=7p=7 is the most classical one. One needs to find a group of order 7, and one such subgroup is given by the group of upper triangular matrices (101)\begin{pmatrix} 1 & * \\ 0 & 1\end{pmatrix}. What makes things work is that pp is the characteristic of the chosen finite field. In general, if qq is a power of pp, then the subgroup of upper-triangular matrices in GL(n,Fq)\mathop{\rm GL}(n,\mathbf F_q) with 11s one the diagonal has cardinality qq2qn1=qn(n1)/2q\cdot q^2\cdots q^{n-1}=q^{n(n-1)/2}, which is exactly the highest power of pp divising the cardinality of GL(n,Fq)\mathop{\rm GL}(n,\mathbf F_q).

Let's now study p=3p=3. We need to find a group SS of order 32=93^2=9 inside GG. There are a priori two possibilities, either S(Z/3Z)2S\simeq (\mathbf Z/3\mathbf Z)^2, or S(Z/9Z)S\simeq (\mathbf Z/9\mathbf Z).
We will find a group of the first sort, which will that the second case doesn't happen, because all 33-Sylows are pairwise conjugate, hence isomorphic.

Now, the multiplicative group F7×\mathbf F_7^\times is of order 66, and is cyclic, hence contains a subgroup of order 33, namely C={1,2,4}C=\{1,2,4\}. Consequently, the group of diagonal matrices with coefficients in CC is isomorphic to (Z/3Z)2(\mathbf Z/3\mathbf Z)^2 and is our desired 33-Sylow.

Another reason why GG does not contain a subgroup SS isomorphic to Z/9Z\mathbf Z/9\mathbf Z is that it does not contain elements of order 99. Let's argue by contradiction and consider a matrix AGA\in G such that A9=IA^9=I; then its minimal polynomial PP divides T91T^9-1. Since 797\nmid 9, the matrix AA is diagonalizable over the algebraic closure of F7\mathbf F_7. The eigenvalues of AA are eigenvalues are 99th roots of unity, and are quadratic over F7\mathbf F_7 since deg(P)2\deg(P)\leq 2. On the other hand, if α\alpha is a 99th root of unity belonging to F49\mathbf F_{49}, one has α9=α48=1\alpha^9=\alpha^{48}=1, hence α3=1\alpha^3=1 since gcd(9,48)=3\gcd(9,48)=3. Consequently, α\alpha is a cubic root of unity and A3=1A^3=1, showing that AA has order 33.

It remains to treat the case p=2p=2, which I find slightly trickier. Let's try to find elements AA in GG whose order divides 252^5. As above, it is diagonalizable in an algebraic closure, its minimal polynomial divides T321T^{32}-1, and its roots belong to F49\mathbf F_{49}, hence satisfy α32=α48=1\alpha^{32}=\alpha^{48}=1, hence α16=1\alpha^{16}=1. Conversely, F49×\mathbf F_{49}^\times is cyclic of order 4848, hence contains an element of order 1616, and such an element is quadratic over F7\mathbf F_7, hence its minimal polynomial PP has degree 22. The corresponding companion matrix AA in GG is an element of order 1616, generating a subgroup S1S_1 of GG isomorphic to Z/16Z\mathbf Z/16\mathbf Z. We also observe that α8=1\alpha^8=-1 (because its square is 11); since A8A^8 is diagonalizable in an algebraic closure with 1-1 as the only eigenvalue, this shows A8=IA^8=-I.

Now, there exists a 22-Sylow subgroup containing S1S_1, and S1S_1 will be a normal subgroup of SS (because its index is the smallest prime number dividing the order of SS, which is 22). This suggests to introduce the normalizer NN of S1S_1 in GG. One then has S1SNS_1\subset S\subset N. Let sSs\in S be such that s∉S1s\not\in S_1; then there exists a unique k{1,,15}k\in\{1,\dots,15\} such that s1As=Aks^{-1}As=A^k, and s2As2=Ak2=As^{-2}As^2=A^{k^2}=A (because ss has order 22 modulo S1S_1), hence k21(mod16)k^2\equiv 1\pmod{16}—in other words, k±1(mod8)k\equiv \pm1\pmod 8.

There exists a natural choice of ss: the involution (s2=Is^2=I) which exchanges the two eigenspaces of AA. To finish the computation, it's useful to take a specific example of polynomial PP of degree 22 whose roots in F49\mathbf F_{49} are primitive 1616th roots of unity. In other words, we need to factor the 1616th cyclotomic polynomial Φ16=T8+1\Phi_{16}=T^8+1 over F7\mathbf F_7 and find a factor of degree 22; actually, Galois theory shows that all factors have the same degree, so that there should be 4 factors of degree 22.  To explain the following computation, some remark is useful. Let α\alpha be a 1616th root of unity in F49\mathbf F_{49}; we have (α8)2=1(\alpha^8)^2=1 but α81\alpha^8\neq 1, hence α8=1\alpha^8=-1.  If PP is the minimal polynomial of α\alpha, the other root is α7\alpha^7, hence the constant term of PP is equal to αα7=α8=1\alpha\cdot \alpha^7=\alpha^8=-1.

We start from T8+1=(T4+1)22T4T^8+1=(T^4+1)^2-2T^4 and observe that 242(mod7)2\equiv 4^2\pmod 7, so that T8+1=(T4+1)242T4=(T4+4T2+1)(T44T2+1)T^8+1=(T^4+1)^2-4^2T^4=(T^4+4T^2+1)(T^4-4T^2+1). To find the factors of degree 22, we remember that their constant terms should be equal to 1-1. We thus go on differently, writing T4+4T2+1=(T2+aT1)(T2aT1)T^4+4T^2+1=(T^2+aT-1)(T^2-aT-1) and solving for aa: this gives 2a2=4-2-a^2=4, hence a2=6=1a^2=-6=1 and a=±1a=\pm1. The other factors are found similarly and we get
T8+1=(T2T1)(T2+T1)(T24T1)(T2+4T1). T^8+1=(T^2-T-1)(T^2+T-1)(T^2-4T-1)(T^2+4T-1).
We thus choose the factor T2T1T^2-T-1 and set A=(0111)A=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}.

Two eigenvectors for AA are v=(1α)v=\begin{pmatrix} 1 \\ \alpha \end{pmatrix} and v=(1α)v'=\begin{pmatrix}1 \\ \alpha'\end{pmatrix}, where α=α7\alpha'=\alpha^7 is the other root of T2T1T^2-T-1. The equations for BB are Bv=vBv=v' and Bv=vBv'=v; this gives B=(1011)B=\begin{pmatrix} 1 & 0 \\ 1 & - 1\end{pmatrix}. The subgroup S=A,BS=\langle A,B\rangle generated by AA and BB has order 3232 and is a 22-Sylow subgroup of GG.

Generalizing this method involves finding large commutative pp-subgroups (such as S1S_1) which belong to appropriate (possibly non-split) tori of GL(n)\mathop{\rm GL}(n) and combining them with adequate parts of their normalizer, which is close to considering Sylow subgroups of the symmetric group. The paper Sylow pp-subgroups of the classical groups over finite fields with characteristic prime to pp by A.J. Weir gives the general description (as well as for orthogonal and symplectic groups), building on an earlier paper in which he constructed Sylow subgroups of symmetric groups. See also the paper Some remarks on Sylow subgroups of the general linear groups by C. R. Leedham-Green and W. Plesken which says a lot about maximal pp-subgroups of the general linear group (over non-necessarily finite fields). Also, the question was recently the subject of interesting discussions on MathOverflow.

[Edited on Febr. 14 to correct the computation of the 2-Sylow...]