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.

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, $G=\mathop{\rm GL}(2,\mathbf F_7)$. 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 $G$. The first column of a matrix in $G$ must be non-zero, hence there are $7^2-1$ possibilities; for the second column, it only needs to be non-collinear to the first one, and each choice of the first column forbids $7$ second columns, hence $7^2-7$ possibilities. In the end, one has $\mathop{\rm Card}(G)=(7^2-1)(7^2-7)=48\cdot 42=2016$. The same argument shows that the cardinality of the group $\mathop{\rm GL}(n,\mathbf F_q)$ is equal to $(q^n-1)(q^n-q)\cdots (q^n-q^{n-1})=q^{n(n-1)/2}(q-1)(q^2-1)\cdots (q^n-1)$.

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

The cas $p=7$ 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 $\begin{pmatrix} 1 & * \\ 0 & 1\end{pmatrix}$. What makes things work is that $p$ is the characteristic of the chosen finite field. In general, if $q$ is a power of $p$, then the subgroup of upper-triangular matrices in $\mathop{\rm GL}(n,\mathbf F_q)$ with $1$s one the diagonal has cardinality $q\cdot q^2\cdots q^{n-1}=q^{n(n-1)/2}$, which is exactly the highest power of $p$ divising the cardinality of $\mathop{\rm GL}(n,\mathbf F_q)$.

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

Now, the multiplicative group $\mathbf F_7^\times$ is of order $6$, and is cyclic, hence contains a subgroup of order $3$, namely $C=\{1,2,4\}$. Consequently, the group of diagonal matrices with coefficients in $C$ is isomorphic to $(\mathbf Z/3\mathbf Z)^2$ and is our desired $3$-Sylow.

Another reason why $G$ does not contain a subgroup $S$ isomorphic to $\mathbf Z/9\mathbf Z$ is that it does not contain elements of order $9$. Let's argue by contradiction and consider a matrix $A\in G$ such that $A^9=I$; then its minimal polynomial $P$ divides $T^9-1$. Since $7\nmid 9$, the matrix $A$ is diagonalizable over the algebraic closure of $\mathbf F_7$. The eigenvalues of $A$ are eigenvalues are $9$th roots of unity, and are quadratic over $\mathbf F_7$ since $\deg(P)\leq 2$. On the other hand, if $\alpha$ is a $9$th root of unity belonging to $\mathbf F_{49}$, one has $\alpha^9=\alpha^{48}=1$, hence $\alpha^3=1$ since $\gcd(9,48)=3$. Consequently, $\alpha$ is a cubic root of unity and $A^3=1$, showing that $A$ has order $3$.

It remains to treat the case $p=2$, which I find slightly trickier. Let's try to find elements $A$ in $G$ whose order divides $2^5$. As above, it is diagonalizable in an algebraic closure, its minimal polynomial divides $T^{32}-1$, and its roots belong to $\mathbf F_{49}$, hence satisfy $\alpha^{32}=\alpha^{48}=1$, hence $\alpha^{16}=1$. Conversely, $\mathbf F_{49}^\times$ is cyclic of order $48$, hence contains an element of order $16$, and such an element is quadratic over $\mathbf F_7$, hence its minimal polynomial $P$ has degree $2$. The corresponding companion matrix $A$ in $G$ is an element of order $16$, generating a subgroup $S_1$ of $G$ isomorphic to $\mathbf Z/16\mathbf Z$. We also observe that $\alpha^8=-1$ (because its square is $1$); since $A^8$ is diagonalizable in an algebraic closure with $-1$ as the only eigenvalue, this shows $A^8=-I$.

Now, there exists a $2$-Sylow subgroup containing $S_1$, and $S_1$ will be a normal subgroup of $S$ (because its index is the smallest prime number dividing the order of $S$, which is $2$). This suggests to introduce the normalizer $N$ of $S_1$ in $G$. One then has $S_1\subset S\subset N$. Let $s\in S$ be such that $s\not\in S_1$; then there exists a unique $k\in\{1,\dots,15\}$ such that $s^{-1}As=A^k$, and $s^{-2}As^2=A^{k^2}=A$ (because $s$ has order $2$ modulo $S_1$), hence $k^2\equiv 1\pmod{16}$—in other words, $k\equiv \pm1\pmod 8$.

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

We start from $T^8+1=(T^4+1)^2-2T^4$ and observe that $2\equiv 4^2\pmod 7$, so that $T^8+1=(T^4+1)^2-4^2T^4=(T^4+4T^2+1)(T^4-4T^2+1)$. To find the factors of degree $2$, we remember that their constant terms should be equal to $-1$. We thus go on differently, writing $T^4+4T^2+1=(T^2+aT-1)(T^2-aT-1)$ and solving for $a$: this gives $-2-a^2=4$, hence $a^2=-6=1$ and $a=\pm1$. The other factors are found similarly and we get
\[ T^8+1=(T^2-T-1)(T^2+T-1)(T^2-4T-1)(T^2+4T-1). \]
We thus choose the factor $T^2-T-1$ and set $A=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$.

Two eigenvectors for $A$ are $v=\begin{pmatrix} 1 \\ \alpha \end{pmatrix}$ and $v'=\begin{pmatrix}1 \\ \alpha'\end{pmatrix}$, where $\alpha'=\alpha^7$ is the other root of $T^2-T-1$. The equations for $B$ are $Bv=v'$ and $Bv'=v$; this gives $B=\begin{pmatrix} 1 & 0 \\ 1 & - 1\end{pmatrix}$. The subgroup $S=\langle A,B\rangle$ generated by $A$ and $B$ has order $32$ and is a $2$-Sylow subgroup of $G$.

Generalizing this method involves finding large commutative $p$-subgroups (such as $S_1$) which belong to appropriate (possibly non-split) tori of $\mathop{\rm GL}(n)$ and combining them with adequate parts of their normalizer, which is close to considering Sylow subgroups of the symmetric group. The paper Sylow $p$-subgroups of the classical groups over finite fields with characteristic prime to $p$ 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 $p$-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...]