Central Extensions and Weight One Forms

As mentioned in the comments to the last post, Kevin Buzzard and Alan Lauder have made an extensive computation of weight one modular forms in characteristic zero (see also here). Thinking about what that data might contain, I wondered about the following question: what are the images of the Galois representations associated to the weight one forms of type A_5?

Let us take a step back. Consider a projective representation

\psi: G_{{\mathbf{Q}}} \rightarrow {\mathrm{PGL}}_2({\mathbf{C}})

with image A_5, and assume that it is odd. (That is, complex conjugation has order 2.) According to Tate, there exists a lift

\rho: G_{{\mathbf{Q}}} \rightarrow {\mathrm{GL}}_2({\mathbf{C}}).

This lift is unique up to twisting. Since the Schur multiplier H_2(A_5,{\mathbf{Z}}) of A_5 is {\mathbf{Z}}/2{\mathbf{Z}}, there is a unique minimal lift up to twist whose image is a central extension {\widetilde{A}_5} by a cyclic group \Delta of 2-power order. Note that \Delta is not trivial, since A_5 does not have any two-dimensional representations. If |\Delta| = 2, then the determinant of the corresponding 2-dimensional representation of {\widetilde{A}_5} is trivial, which contradicts the assumption that \psi is odd. (Equivalently, there is an obstruction at \infty to lifting to the central extension by {\mathbf{Z}}/2{\mathbf{Z}}.) Hence 4 divides |\Delta|. What is the expected distribution of \Delta as one runs over all odd A_5-extensions?

My first guess (without any prior thought or computation) was that this might obey some form of Cohen–Lenstra heuristic, suitably interpreted.

Note that the image of the determinant has order |\Delta|/2. The corresponding determinant representation is a character of {\mathbf{Q}} of 2-power order. Since {\mathbf{Q}} has trivial class number, the order |\Delta|/2 is equal to the maximal ramification degree e_p of this representation over all primes p.

Over {\mathbf{Q}}, Tate’s lifting theorem has the following stronger form: one may choose a lift \rho_p of \psi_p := \psi|_{D_p} and insist that \rho|_{I_p} = \rho_p | I_p; that is, they agree on inertia. This is essentially a consequence of the fact that {\mathbf{Q}} has trivial class group. For convenience, suppose that \psi is unramified at 2 and 3. Suppose that \psi is ramified at p. There are three possibilities:

  1. The image of \psi_p at a ramified prime p is cyclic of order 2, 3, or 5.
  2. The image of \psi_p at a ramified prime p is S_3.
  3. The image of \psi_p at a ramified prime p is {\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2{\mathbf{Z}}.

For a fixed p, let \epsilon denote the Teichmuller lift of the mod-p cyclotomic character. (Fix an isomorphism of \mathbf{C} with \overline{\mathbf{Q}}_p for all p.)

Let us consider the three cases in turn.

In the first case, the image factors through a cyclic quotient. One may thus take \rho_p to be a direct sum which, on inertia, has the shape \chi \oplus 1 up to twist. By comparing this to the projective representation, we see that \chi has order 2, 3, or 5, and so, after finding the twist such that the determinant has 2-power order, we see that e_p = 1 or e_p = 2.

In the second case, the lift on inertia is (up to twist) of the form:

\omega^{r}_2 \oplus \omega^{pr}_2

for some r. Since the order of \omega_2 is p^2 - 1, the order of the ratio is


which must be equal to 3. It follows that r is even. Yet the determinant is equal to

\displaystyle{ \omega^{(p+1)r}_2 = \epsilon^{r}},

Since r is even, we see that, after twisting, we may take e_p = 1.

Finally, in the third case, the lift is of the form:

\omega^{r}_2 \oplus \omega^{pr}_2

for some r. We now find that the order of the ratio of these characters is


which must be equal to 2, and the determinant is \epsilon^r. If r is even, then, as above, we may twist so that e_p = 1. Hence, the only way that the image after minimal twist does not have |\Delta| = 4 is if we are in this third situation with p \equiv 1 \mod 4, with r odd, and then (after twisting) we find that e_p is the largest power of 2 dividing p - 1.

(I confess that I originally forgot the fact that the third possibility could occur, and was only after noticing that this seemed to imply the inverse Galois problem was false thought a little bit more about the possibilities.)

To summarize:

Lemma Assume that \psi is unramified at 2 and 3 and has projective image A_5, and a lift with image \widetilde{A}_5 with minimal kernel. Then order of \Delta is 4 unless there exists a prime p \equiv 1 \mod 4 such that the image of the decomposition group at p under \psi is {\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}. In this case, we have \Delta to be twice the largest power of 2 dividing p -1 for all such primes p.

Let \Delta(\psi) denote the corresponding power of 2.

We see that \Delta(\psi) is determined by purely local phenomena. This still doesn’t quite answer what the distribution of the extension \widetilde{A}_5 will be. However, I imagine that Bhargava style heuristics should certainly be able to predict the ratio of A_5 with \Delta(\psi) = 2^n. Does anyone have a sense of how easy this might be to prove? Or, much more modestly, how easy it would be to compute from these quite precise heuristics the exact predicted distribution of the central extensions of A_5 coming from weight one modular forms?

(I confess, it is not even obvious to me from this construction how to prove that all central extensions \widetilde{A}_5 occur as Galois groups — but I presume this is known, and hopefully one of my readers can provide a reference.)

(According to KB, BTW, all the A_5 representations with N \le 1500 have \Delta = 4.)

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The LMFDB has gone live!

I previously expressed on this blog a somewhat muted opinion about certain aspects of the website’s functionality, and it seems that my complaints have mainly been addressed in the latest version. On the other hand, this live version has been released with a certain amount of fanfare, not to mention press releases. (Is AIM involved? yes it is.) First of all, there’s something about press releases in science (or mathematics) which I find deeply troubling. What is the point of such a press release? To drum up future funding? To generate mainstream press articles and thus communicate a sense of wonder and amazement to the public, who rarely get to glimpse the excitement of modern mathematics in action? Hopefully it’s the later which is true. But if so, does it really require that we stretch the truth about what we are doing in order to generate such excitement? (To see that the answer to the latter question is no, one need only look back on Scientific American articles on mathematics and physics from the ‘60s and ‘70s.) To be fair, I should also link to a more modest description of the project here. But then again, I dare you to click on the following website.

But back to the topic at hand. I have asked the opinion of at least three mathematicians about either the LMFDB or on computational aspects of the Langlands Programme more generally. For reasons of anonymity, I will not mention their names here. The first comment addresses a widely circulated quote from John Voight in the press release: “Our project is akin to the first periodic table of elements.” One source offered the following take on this (literally copied from my email and modified only by adjusting the spelling of the Langlands Programme to the preferred Canadian spelling):

The periodic table was a fantastic synthesis of decades, maybe even centuries, of empirical investigation, that led to profound new theoretical insights into chemistry and the physical world. The LMFDB is a rag-tag assortment of empirical facts in the Langlands Programme which lag far behind the theoretical advances of the past decade.

A second source, speaking more broadly on the question of computations in the Langlands programme, wondered if there had been any fundamental discoveries made via numerical computations since the BSD conjecture. While these are certainly pretty strong statements, I feel comfortable agreeing at least that, in our field, the theory is way in advance of the computations. We can prove potential modularity theorems for self-dual representations of any dimension, and yet it’s barely possible to compute even weight zero forms for U(3) (David Loeffler did some computations once). In part, I think this actually provides justification for effort into understanding how to actually compute these objects. After all, a really nice aspect of number theory is having a collection of beautiful concrete examples, from X_0(11) to Q(sqrt(-23)) (take that, Geometric Langlands!). Yet these statements also provide an alternate framework with which to view the LMFDB, which is less glorious than the press releases suggest. To continue with the periodic table analogy, the LMDFB is less a construction of a periodic table, but more a collection of sample elements from the periodic table neatly contained in small glass vials. Let me make the following point clear: some of the samples took a great deal of effort and ingenuity to extract. But it’s not entirely obvious to me how easy it will be to actually use the data to do fundamental new mathematics. This was the main point of my third commentator: the problem is that, if one actually wants to undertake a fairly serious computation, the complete set of data one has to compute will either not be available on the LMFDB (however big the LMDFB grows), or not available in a format which is at all practical to extract for actually doing computations. So, in the end, if you need some serious data, you are probably going to have to compute it again yourself. In some cases you may be able to do this, and in other cases not.

This leads to probably the most frustrating thing about the website from my perspective. I thought quite a bit about what the most useful format for some of the data might be (for me). Here is a typical thing that I might want to do: find a Hecke eigenform of some particular weight and level, and then compute information about the mod-p representations for various primes p (as well as congruences between forms). This is a little tricky to do at the moment, in part because the data for the coefficients is given in terms of a primitive element in the Hecke field (often the eigenvalue of T_2), and then the order generated by this element has (almost always) huge index (divisible by many small primes) inside the ring of integers, which makes computing the reductions slightly painful. There are certainly ways to address this specific problem and incorporate such functionality into the website. But it ultimately would be silly to customize the LMFDB for my particular needs. Instead, in the end, what I think would actually be most useful would be if the webpages on modular forms included enough pari/gp/sage/magma code to allow me to go away and compute the q-expansion myself. This is why I think that, even within computational number theory, the impact of programs like pari/sage/mamga will be far greater than the LMFDB.

If I think of the three most serious computations I have been involved with recently, they include partial weight one Hilbert modular forms, non-liftable classes of low weight Siegel modular forms, and Artin L-functions of S_5 extensions. The first required customized programs in magma (some written in part by John Voight), and the tables of higher weight HMFs in the LMFDB would not have been of any use. The computations of low weight Siegel modular forms in finite characteristic, which are absolutely terrifying, require completely custom computations (which, by the way, are completely beyond my capability of doing and involve getting Cris Poor and David Yuen to do them). Finally, the computation which would be the closest to an off the shelf computation was proving that the Artin L function L(\rho_5,s) of some S_5 extension provably had no zero in [0,1]. However, the LMFDB tables only go up to degree four representations. Fortunately I knew that Andy Booker was an expert in this sort of thing and he did it for me. (Then again, even if the data in this case was included, it’s totally unclear to me the extent to which the data in the tables has been “proved.”) And my point here is not to complain about the lack of certain computations being in the LMFDB, just to caution against the idea that its existence will be particularly useful for future computations.

The bigger point, however, is surely this: Is hype in science or mathematics a necessary evil to generate public enthusiasm, or is it an ultimately corrosive influence?

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Report From Berkeley

My recent trip to Berkeley did not result in a chance to test whether the Cheeseboard pizza maintained its ranking, but did give me the opportunity to attend the latest Bay Area Number Theory and Algebraic Geometry day, on a (somewhat disappointingly) rainy Saturday in Evans Hall. The weather was somewhat better on Sunday, however, allowing myself to make the trip to Mint Plaza for the following cup, which should bear some resemblance to the banner picture on this site. (Unfortunately, they were no longer serving their mini-Brioche buns.)

Blue Bottle Coffee

But now on to the good stuff, a report on some of the talks:

Jaclyn Lang gave a talk on her work concerning the image of big Galois representions. The setup is roughly as follows. Let

\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)

be an absolutely irreducible odd Galois representation over a finite field (hence modular). Suppose this Galois representation was the residual representation associated to an ordinary modular form that lived inside a Hida family that was smooth over weight space. Then one might expect that the corresponding representation

\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{Z}_p[[T]])

to have image which was as big as possible, namely, containing \mathrm{SL}_2(\mathbf{Z}_p[[T]]). This can’t always be the case, however; for example, the residual representation (or the entire family) could be dihedral. However, if the residual representation contains \mathrm{SL}_2(\mathbf{F}_p), and one additionally assumes that the image of inertia at p is sufficiently large, then this is indeed the case (probably this assumes that the residue characteristic is at least 5). There have been a number of generalizations of this result due to Hida and others which improves the result by weaken the various hypotheses; for example, allowing coefficients, allowing the residual representation to be dihedral, and weakening the ramification hypothesis at p. For these results, one can’t expect that the image is full, but rather that the image contains an appropriate congruence subgroup of \mathrm{SL}_2. I like to think of this as follows: at classical specializations, one knows that the image (if it is not CM or weight one) will have open image; the results of this talk and of previous work show that index can be controlled in families. Actually, this is not quite true, because another obstruction to having open image even classically is the existence of inner twists. The main result of the talk was to deal with this issue of inner twists, and hence also allow for a generalization of the results not only to smooth Hida families but to any irreducible component of any Hida family. (More details to be found here.)

A natural question: one output of Lang’s result is to give an ideal \mathfrak{b} of the Hida family for which the image of these Galois representations contains the \mathfrak{b}-congruence subgroup (after accounting for inner twists). In characteristic zero, my impression from the talk was that one can identify the support of this ideal as coming from CM points and classical weight one modular forms. On the other hand, apparently there is also a version of this result in the reducible case (due to Hida and with extra hypotheses); in that case the zeros should correspond to the reducible locus, or equivalently, the zeroes of the p-adic zeta function. However, a stronger result is true, namely, that \mathfrak{b} can essentially be identified with this p-adic zeta function. So, returning back to the residually irreducible case, the natural question is: can the support of \mathfrak{b} contain the prime p?

Kestutis Cesnavicius gave a talk on the Manin-Stevens and Manin constants for elliptic curves, with emphasis on the prime p=2. He raised the following question: Suppose that N is odd. Is there a surjection from the space of weight 2 classical modular cusp forms of level \Gamma_0(N) with coefficients in \mathbf{Z}_2 to the space of weight 2 Katz cusp forms of the same level over \mathbf{F}_2? The issue here is that the latter space is really the cohomology of the associated stack, not the course moduli space. Unfortunately, this question distracted me a little as I tried to find a counter-example (I failed). A result of Serre and Carayol basically implies that the result can only fail after localizing at a non-Eisenstein maximal ideal \mathfrak{m} of the Hecke algebra \mathbf{T} if the corresponding representation \overline{\rho}_{\mathfrak{m}} is induced from \mathbf{Q}(\sqrt{-1}). (Analogously, for p =3, when the representation is induced from \mathbf{Q}(\sqrt{-3}).) This is related to the classic failure of the first version of Serre’s conjecture for p =3 at level \Gamma_1(13). However, as Serre quickly realized, this failure ultimately comes from a failure to lift mod-p forms as above, except in this case from the intermediate curve X_H(13), not from X_0(13). I ultimately convinced myself that lifting was always possible unless \mathfrak{m} was not only Eisenstein but also the ideal containing T_{p} for all odd p not dividing N. I think this must be related to Ken’s result on component groups of Neron models, and how the non-Eisenstein parts arise for X_H(N) but not for X_0(N) or X_1(N). (More details here.)

The final speaker of the day was Daquin Wan. The key question that arose in his talk was the following. Suppose that D(k,T) is the characteristic power series of the U operator on the space of overconvergent p-adic modular forms in integral weight k. Can one show that

L(k,T) = \displaystyle{\frac{D(k+2,T)}{D(k,pT)}}

has infinitely many zeros and infinitely many poles? One actually has to assume that k \ne 0 here, since otherwise the result is false, as this will be a polynomial of dimension the space of weight two forms. One feels that p-adic Langlands should be able to say enough about slopes in these weights to obtain a contradiction, but I don’t unfortunately see how to do it. The main point of the talk was two-fold. There is an argument of Coleman that shows that D(k,T) is not itself a polynomial. This argument can be generalized to prove that L(k,T) is not a rational function. Second, the product L(k,T) L(-k,p^k T) is actually a rational function because of the properties of the theta operator. So one deduces that at least one of these functions had infinitely many poles and the other had infinitely many zeroes. This also relies on a previous result of Wan that these functions are meromorphic. (Oh, I should mention that this was joint work with … and here I didn’t take notes for a talk two weeks ago … Liang Xiao? Please correct me if I’m wrong)

(Romyar Sharifi also talked, but since I am actively trying to understand something about that talk on a more technical level, so I will have to return to a discussion of it later.)

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Tensor Products

Let W be an irreducible representation of a finite group G. Say that W is tensor indecomposable if any isomorphism W = U \otimes V implies that either U or V is a character.
In conversations with Matt and Toby (which permeate the rest of this post as well), the following problem came up:

Question: Let G be a finite group. Let V be an irreducible representation of G. Is there a unique decomposition

V = V_1 \otimes V_2 \ldots \otimes V_k

of V as a tensor product of tensor indecomposable representations (up to re-ordering and twist)? I don’t think this can be too hard, but I confess I don’t see how to do it. (Since we didn’t really need this, we didn’t think about it too hard.)
(edit: when I say I don’t think this can be too hard, I don’t mean to imply that I think it is true; just that I think either a proof or counterexample should not be too hard to find — hopefully not both.)

One can ask an analogous problem for Lie groups. Actually, the problem for Lie algebras is actually quite simple (and the answer is positive). It is related to the following:

Let V and W be irreducible non-trivial representations of a simple Lie group \mathfrak{g}. Then V \otimes W is reducible.

Proof: Assume that V \otimes W is irreducible. In particular, it is determined by its highest weight. Let the highest weights of V and W be \lambda and \mu respectively. Then the highest weight of V \otimes W must be \lambda + \mu. But now, by the Weyl character formula, we deduce that

\displaystyle{1 = \frac{\dim(V) \dim(W)}{\dim(V \otimes W)} = \prod_{\Phi^{+}} \frac{ \langle \rho,\alpha \rangle \langle \rho + \lambda + \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

The product term can also be written as:

\displaystyle{1 + \frac{ \langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.

In particular, since the pairing is non-negative between positive roots and highest weights, we deduce a contradiction unless

\langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle = 0

for all \alpha \in \Phi^{+}. The assumption that \mathfrak{g} is irreducible, however, is equivalent to saying that \Phi^{+} has a maximal root \beta, and for such a maximal root, we have

\langle \lambda, \beta \rangle = 0 \Rightarrow \langle \lambda, \alpha \rangle = 0, \ \forall \alpha \in \Phi^{+} \Leftrightarrow \lambda = 0.

Note that this lemma is actually a special case of a theorem of Rajan, who proved that, for simple \mathfrak{g}, the factors of a (not necessarily irreducible) tensor product are determined by the representation. In particular, the tensor product of two non-trivial irreducible representations cannot be irreducible.

The problem with the initial question is that it’s hard to construct tensor products of irreducible representations which are irreducible. Or rather, it is easy, simply by taking U \otimes V where U is an irreducible representation of H and V is an irreducible representation of G and the tensor is irreducible for H \times G. Yet the interesting case is something closer to assuming that U and V are faithful. Actually, this motivates the following question:

Question: Do there exist irreducible non-trivial representations U and V of a finite simple non-abelian group G such that U \otimes V is irreducible?

The argument above for Lie groups suggests that this may not happen for Chevelley groups (although it certainly doesn’t prove this). It also suggests (relating the representation theory of A_n and S_n to \mathrm{GL}_n) that it doesn’t happen for the alternating groups either. It almost surely doesn’t happen for the sporadic groups either. So my guess that the answer to the problem above is no, and that this is probably known, and probably requires classification. (Please comment if you know the answer.) Actually, this also reminds me of a similar problem which I think is open.

Question: Fix N. Does there exist a non-trivial representation V of a finite group G of dimension N such that \mathrm{Hom}^0(V,V) (of dimension N^2 - 1) is irreducible?

This question came up in my paper with Barry, where I was surprised to find very few examples. I seem to remember that the Mathieu group M_{12} has an 11-dimensional representation whose corresponding 120 dimensional adjoint is irreducible. Can one classify all such examples coming from simple groups?

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Due Process

When I was a senior (’93), my high school began a program of inviting “distinguished” visitors to give talks to a selected number of students, several of whom were also invited to lunch with the speaker. I was fortunate that the program was run by Lawrence (Larry) Doolan, head of maths, which meant that I got more than my fair share of free lunches. I’m not sure how many talks there were, but I particularly remember talks by John Halfpenny (who gave, perhaps unexpectantly, a thoughful and nuanced take on labour unions), Meg Lees (ugh, no wonder the Democrats no longer exist as a party), Michael Dukakis, and also someone who I believe was a judge on the International Court of Justice. I remember at the time (over lunch) questioning the latter gentleman what the ICJ (and UN more generally) was planning on doing about the ongoing crisis in Yugoslavia. He responded (somewhat peevishly in my memory) that this was an internal matter which was not within the juristriction of any UN courts. To my young mind, he certainly seemed to display a perverse attitude which elevated process far above justice. It was therefore satisfying today to see that the International Criminal Court (which did not exist until 2002) sentence Radovan Karadzic to 40 years in prison. I mean, it’s cetainly at least 20 years too late, but better than nothing. (Slobodan Milošević, on the other hand, got off lightly.)

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Ventotene, Part II

I promised to return to a more mathematical summary of the conference in Ventotene, and indeed I shall do so in the next two posts.

One of the themes of the conference was bounding the order of the torsion subgroup in arithmetic lattices. Tsachik Gelander gave a number of talks (in part) on the seven author paper. One nice result was a uniform bound of the shape

\log |H^*(\Gamma,\mathbf{Z})^{\mathrm{tors}}| < C \cdot \mathrm{Vol}(\Gamma),

where \Gamma ranges (say) over all lattices in \mathrm{SL}_n(\mathbf{R}) for a fixed n \ge 3. (The key result here is the uniformity — this result is much easier for covers of a fixed manifold.)

Two natural questions that came up (in conversation at least) during the conference are as follows:

  1. Can one do better in low degree?
  2. What is the true expectation for the size of this group for (say) congruence subgroups of \mathrm{SL}_n(\mathbf{Z})?

Let’s consider the first question. For (congruence) subgroups of \mathrm{SL}_n(\mathbf{Z}), one can certainly say quite a bit more. For example, H^1 is essentially trivial, by the congruence subgroup property. However, in the stable range of cohomology (in particular, when the completed cohomology groups become stable), the groups H^* are finite over \mathbf{Z}_p, and so contribute very little. One does, at least, have the following soft arguments for general groups.

Proposition: Let G be a semi-simple group over \mathbf{Q} with \mathbf{Q}-rank r = r_{\mathbf{Q}}. Then \widetilde{H}_i is a torsion \Lambda = \mathbf{Z}_p[[G(\mathbf{Z_p})]]-module for i <  r_{\mathbf{Q}}.

The proof is as follows: the boundary terms are also torsion, so it suffices to show that all the \widetilde{H}^{BM}_i in the appropriate range are also torsion, where we consider Borel-Moore homology. Assume otherwise. Let \dim G(\mathbf{R})/K(\mathbf{R}) = d. From the spectral sequence \mathrm{Ext}^i(\widetilde{H}^{BM}_j,\Lambda) \Rightarrow \widetilde{H}_{d-i-j}, we deduce that there is at least one i < r_{\mathbf{Q}} such that \widetilde{H}_{d-i} \ne 0. Yet the homological dimension of \Gamma \backslash G(\mathbf{R})/K(\mathbf{R}) is, by Borel-Serre, equal to d - r_{\mathbf{Q}}, and so all the homology in these degrees (and hence certainly the completed homology) vanishes.

One can do better in certain algebraic cases, where one can deduce vanishing of the completed cohomology in certain degrees by perfectoid technology (as in Corollary 4.2.3 of Scholze’s paper).

The answer to the second question, even conjecturally, is more mysterious. There are some speculations related to this question in Bergeron-Venkatesh. But it seems a little tricky to formulate a precise guess (for a good upper bound).

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Best seat in the house

I recently attended an András Schiff recital at Chicago Symphony Center (not the first such Schiff performance I have been to). The conceit of this concert was the “last sonatas,” a performance of the last sonatas of Beethoven, Haydn, Mozart, and Schubert respectively. (This was actually the third in a series of concerts, the previous two consisting of the antepenultimate and penultimate sonatas respectively of the same four composers.) I was a little bit disorganized about buying tickets — I only heard about the concert on the radio the week before — but this had a positive effect: the tickets were almost sold out, so the CSO introduced extra “stage seating,” which is exactly what it sounds like. I literally had the best seat in the house, roughly located in the middle of where the first violins would be. I swear that Andras Schiff looked directly into my eyes for one of his bows. (To be fair, hew bowed quite a few times — there were about 6 curtain calls in between two encores: a middle movement of the penultimate Schubert sonata and the aria to the Goldberg variations.) If you ever get a chance to purchase stage tickets at the CSO, I strongly recommend it!

Yes, this is really my seat

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