yinhow
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无限维矩阵的行立式
看文献碰到难题，文章中给出了无限维矩阵的元的具体表达式，D(\mu,\nu),\mu,\nu的取值有一部分是整数，有一部分是连续取值，然后文章给出了矩阵的行立式。有无系统的方法来求这种无限维矩阵的行立式？
| 发表时间：2006-05-31, 21:41:03 |
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季候风
发表文章数: 291
武功等级: 太极剑法
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内力值: 370/370
Re: 无限维矩阵的行立式
据我所知, 行列式能扩展到 Fredholm 算子. 计算办法颇为复杂, 参见
Segal's lecture notes: http://www.cgtp.duke.edu/ITP99/segal/
如果能找到 Quillen 关于算子行列式定义的原文当然更好.
还有, 物理学家好像有自己的定义......即使算子的特征值不可加, 也能做 "重整化" 得到有限值
书山有路勤为径
学海无涯苦作舟
| 发表时间：2006-06-01, 00:26:08 |
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kanex
发表文章数: 860
武功等级: 弹指神通
(第六重)
内力值: 343/343
Re: 无限维矩阵的行立式
john baez time!
To make this precise, let's use X to denote a map from the torus to Rn. Then a physical observable will be some function f(X), and its expectation value will be
(1/Z) ∫ f(X) exp(-S(X)) dX
Here S(X) is the action for string theory, which is just the area of the string worldsheet. The number Z is a normalizing factor called the partition function:
Z = ∫ exp(-S(X)) dX
But there's a big problem here! As usual in quantum field thoery, the space we're trying to integrate over is infinite-dimensional, so the above integrals have no obvious meaning. Technically speaking, the problem is that there's no Lebesgue measure "dX" on an infinite-dimensional manifold.
Mathematicians might throw up our hands in despair and give up at this point. But physicists take a more pragmatic attitude: they just keep massaging the problem, breaking rules here and there if necessary, until they get something manageable. Physicists call this "calculating the path integral", but from a certain viewpoint what they're really doing is defining the path integral, since it only has a precise meaning after they're done.
In the case at hand, it was Polyakov who figured out the right massage:
6) A. M. Polyakov, Quantum geometry of bosonic strings, Phys. Lett. B103 (1981), 207.
He rewrote the above integral as a double integral: first an integral over the space of metrics g on the torus, and then inside, for each metric, an integral over maps X from the torus into spacetime. Of course, any such map gives a metric on the torus, so this double integral is sort of redundant. However, introducing this redundancy turns out not to hurt. In fact, it helps!
To keep life simple, let's just talk about the partition function
Z = ∫ exp(-S(X)) dX
If we can handle this, surely we can handle the integral with the observable f(X) in it. Polyakov's trick turns the partition function into a double integral:
Z = ∫ ( ∫ exp(-<X, ΔX>) dX) dg
where Δ is the Laplacian on the torus and the angle brackets stand for the usual inner product of Rn-valued functions, both defined using the metric g.
At first glance Polyakov's trick may seem like a step backwards: now we have two ill-defined integrals instead of one! However, it's actually a step forward. Now we can do the inside integral by copying the formula for the integral of a Gaussian of finitely many variables - a standard trick in quantum field theory. Ignoring an infinite constant that would cancel later anyway, the inside integral works out to be:
(det Δ)-1/2
But wait! The Laplacian here is a linear operator on the vector space of Rn-valued functions on the torus. This is an infinite-dimensional vector space, so we can't blithely talk about determinants the way we can in finite dimensions. In finite dimensions, the determinant of a self-adjoint matrix is the product of its eigenvalues. But the Laplacian has infinitely many eigenvalues, which keep getting bigger and bigger. How do we define the product of all its eigenvalues?
Of course the lowest eigenvalue of the Laplacian is zero, and you might think that would settle it: the product of them all must be zero! But that would make the above expression meaningless, and we are not going to give up so easily. Instead, we will simply ignore the zero eigenvalue! That way, we only have to face the product of all the rest.
(Actually there's something one can do which is slightly more careful than simply ignoring the zero eigenvalue, but I'll talk about that later.)
Okay, so now we have a divergent product to deal with. Well, in "week126" I used a trick called zeta function regularization to make sense of a divergent sum, and we can use that trick here to make sense of our divergent product. Suppose we have a self-adjoint operator A with a discrete spectrum consisting of positive eigenvalues. Then the "zeta function" of A is defined by:
&zeta(s) = tr(A-s)
To compute ζ(s) we just take all the eigenvalues of A, raise them to the -s power, and add them up. For example, if A has eigenvalues 1,2,3,..., then ζ(s) is just the usual Riemann zeta function, which we already talked about in "week126".
This stuff doesn't quite apply if A is the Laplacian on a compact manifold, since one of its eigenvalues is zero. But we have already agreed to throw out the zero eigenvalue, so let's do that when defining ζ(s) as a sum over eigenvalues. Then it turns out that the sum converges when the real part of s is positive and large. Even better, there's a theorem saying that Zeta(s) can be analytically continued to s = 0. Thus we can use the following trick to define the determinant of the Laplacian.
Suppose that A is a self-adjoint matrix with positive eigenvalues. Then
ζ(s) = tr(exp(-s ln A))
Differentiating gives
ζ'(s) = -tr(ln A exp(-s ln A))
and setting s to zero we get
ζ'(0) = -tr(ln A).
But there's a nice little formula saying that
det(A) = exp(tr(ln A))
so we get
det(A) = exp(-ζ'(0)).
Now we can use this formula to define the determinant of the Laplacian on a compact manifold! Sometimes this is called a "regularized determinant".
So - where are we? We used Polyakov's trick to write the partition function of our torus-shaped string as
Z = ∫ ( ∫ exp(-<X, ΔX>) dX) dg,
then we did the inside integral and got this:
Z = ∫ (det Δ)-1/2 dg
and then we figured out a meaning for the determinant here.
What next? Well, since the Laplacian on Rn-valued functions is the direct sum of n copies of the Laplacian on real-valued functions, we expect that
(det Δ)-1/2 = (det laplacian)-n/2
where "laplacian" stands for the Laplacian on ordinary real-valued functions on the torus. One can actually check this rigorously using the definition in terms of zeta functions. That's reassuring: at least one step of our calculation is rigorous! So we get
Z = ∫ (det laplacian)-n/2 dg
Great. But we are not out of the woods yet. We still have an integral over the space of metrics to do - another nasty infinite-dimensional integral.
Time for another massage!
Look at this formula again:
Z = ∫ ( ∫ exp(-<X, ΔX>) dX) dg
The Laplacian depends on the metric g, and so does the inner product. However, the combination <X, ΔX> depends only on the "conformal structure" - i.e., it doesn't change if we multiply the metric by a position-dependent scale factor. It also doesn't change under diffeomorphisms.
Now the space of conformal structures on a torus modulo diffeomorphisms is something familiar: it's just the moduli space of elliptic curves! We figured out what this space looks like in "week125". It's finite-dimensional and there's a nice way to integrate over it, called the Weil-Petersson measure. So we can hope to replace the outside integral - the integral over metrics - by an integral over this moduli space.
Indeed, we could hope that
Z = ∫ (∫ exp(-<X, ΔX>) dX) d[g] [hope!]
Récoltes et semailles
| 发表时间：2006-06-01, 01:18:41 |
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kanex
发表文章数: 860
武功等级: 弹指神通
(第六重)
内力值: 343/343
Re: 无限维矩阵的行立式
where now the outside integral is over moduli space and d[g] is the Weil-Petersson measure. The hope, of course, is that the stuff on the inside is well-defined as a function on moduli space.
Actually this hope is a bit naive. Even though <X, ΔX> doesn't change if we recale the metric, the whole inside integral
∫ exp(-<X, ΔX>) dX
does change. This may seem odd, but remember, we did a lot of hair-raising manipulations before we even got this integral to mean anything! We basically wound up defining it to be
(det Δ)-1/2,
and one can check that this does change when we rescale the metric. This problem is called the "conformal anomaly".
Are we stuck? No! Luckily, there is another problem, which cancels this one when n = 26. They say two wrongs don't make a right, but with anomalies that's often the only way to get things to work....
So what's this other problem? It's that we shouldn't just replace the measure dg by the measure d[g] as I did in my naive formula for the partition function. We need to actually figure out the relation between them. Of course this is hard to do, because the measure dg doesn't exist, rigorously speaking! Still, if we do a bunch more hair-raising heuristic manipulations, which I will spare you, we can get a formula relating dg and and d[g], and using this we get
Z = ∫ ( ∫ exp(-<X, ΔX>) dX) f(g) d[g]
where f(g) is some function of the metric. There's a perfectly explicit formula for this function, but your eyeballs would fall out if I showed it to you. Anyway, the real point is that IN 26 DIMENSIONS AND ONLY IN 26 DIMENSIONS, the integrand
( ∫ exp(-<X, ΔX>) dX) f(g)
is invariant under rescalings of the metric (as well as diffeomorphisms). In other words, the conformal anomaly in
∫ exp(-<X, ΔX>) dX
is precisely canceled by a similar conformal anomaly in f(g), so their product is a well-defined function on moduli space, so it makes sense to integrate it against d[g]. We can then go ahead and figure out the partition function quite explicitly.
By now, if you're a rigorous sort of pure mathematician, you must be suffering from grave doubts about the sanity of this whole procedure. But physicists regard this chain of reasoning, especially the miraculous cancellation of anomalies at the end, as a real triumph. And indeed, it's far better than most of what happens in quantum field theory!
I've heard publishers of science popularizations say that each equation in a book diminishes its readership by a factor of 2. I don't know if this applies to This Week's Finds, but normally I try very hard to keep the equations to a minimum. This week, however, I've been very bad, and if my calculations are correct, by this point I am the only one reading this. So I might as well drop all pretenses of expository prose and write in a way that only I can follow! The real reason I'm writing this, after all, is to see if I understand this stuff.
Okay, so now I'd like to see if I understand how one explicitly calculates this integral:
∫ exp(-<X, laplacian X>) dX
Since we're eventually going to integrate this (times some stuff) over moduli space, we might as well assume the metric on our torus is gotten by curling up the following parallelogram in the complex plane:
τ τ + 1
* *
* *
0 1
There are at least two ways to do the calculation. One is to actually work out the eigenvalues of the Laplacian on this torus and then do the zeta function regularization to compute its determinant. Di Francesco, Mathieu, and Senechal do this in the textbook I talked about in "week124". They get
∫ exp(-<X, laplacian X>) dX = 1 / (√Im(τ) |η(τ)|2)
where "η" is the Dedekind eta function, regarded as function of tau. But the calculation is pretty brutal, and it seems to me that there should be a much easier way to get the answer. The left-hand side is just the partition function for an massless scalar field on the torus, and we basically did that back in "week126". More precisely, we considered just the right-moving modes and we got the following partition function:
1/η(τ)
How about the left-moving modes? Well, I'd guess that their partition function is just the complex conjugate,
1/η(τ)*
since right-movers correspond to holomorphic functions and left-movers correspond to antiholomorphic functions in this Euclidean picture. It's just a guess! And finally, what about the zero-frequency mode? I have no idea. But we should presumably multiply all three partition functions together to get the partition function of the whole system - that's how it usually works. And as you can see, we almost get the answer that Di Francesco, Mathieu, and Senechal got. It would work out perfectly if the partition function of the zero-frequency mode were 1/√Im(τ). By the way, Im(τ) is just the area of the torus.
As evidence that something like this might work, consider this: the zero-frequency mode is presumably related to the zero eigenvalue of the Laplacian. We threw that out when we defined the regularized determinant of the Laplacian, but as I hinted, more careful calculations of
∫ exp(-<X, laplacian X>) dX
don't just ignore the zero eigenvalue. Instead, they somehow use it to get an extra factor of 1/√Im(τ). Admittedly, the calculations are not particularly convincing: a more obvious guess would be that it gives a factor of infinity. Di Francesco, Mathieu, and Senechal practically admit that they need this factor just to get modular invariance, and that they'll do whatever it takes to get it. Nash just sticks in the factor of 1/√Im(τ), mutters something vague, and hurriedly moves on.
Clearly the reason people want this factor is because of how the eta function transforms under modular transformations. In "week125" I said that the group PSL(2,Z) is generated by two elements S and T, and if you look at the formulas there you'll see they act in the following way on tau:
S: τ |→ -1/τ
T: τ |→ τ + 1
The Dedekind eta function satisfies
η(-1/τ) = (τ / i)1/2 η(τ)
η(τ + 1) = exp(2 π i / 24) η(τ)
The second one is really easy to see from the definition; the first one is harder. Anyway, using these facts it's easy to see that
1 / (√Im(τ) |η(τ)|2)
is invariant under PSL(2,Z), so it's really a function on moduli space - but only if that factor of 1/sqrt(Im(tau)) is in there!
Finally, I'd like to say something about why the conformal anomalies cancel in 26 dimensions. When I began thinking about this stuff I was hoping it'd be obvious from the transformation properties of the eta function - since they have that promising number "24" in them - but right now I do not see anything like this going on. Instead, it seems to be something like this: in the partition function
Z = ∫ (∫ exp(-<X, ΔX>) dX) f(g) d[g]
the mysterious function f is basically just the determinant of the Laplacian on vector fields on the torus. So ignoring those darn zero eigenvalues the whole integrand here is
det(laplacian)n/2 det(laplacian')
where "laplacian" is the Laplacian on real-valued functions and " laplacian' " is the Laplacian on vector fields. Now these determinants aren't well-defined functions on the space of conformal structures; they're really sections of certain "determinant bundles". But in this situation, the determinant bundle for the Laplacian on vector fields just so happens to be the 13th tensor power of the determinant bundle for the Laplacian on functions - so the whole expression above is a well-defined function on the space of conformal structures, and thence on moduli space, precisely when n = 26!!!
Now this "just so happens" cannot really be a coincidence. There are no coincidences in mathematics. That's why it pays to be paranoid when you're a mathematician: nothing is random, everything fits into a grand pattern, it's all just staring you in the face if you'd only notice it. (Chaitin has convincingly argued otherwise using Goedel's theorem, and certainly some patterns in mathematics seem "purely accidental", but right now I'm just waxing rhapsodic, expressing a feeling one sometimes gets....)
Indeed, look at the proof in Nash's book that one of these determinant bundles is the 13th tensor power of the other - I think this result is due to Mumford, but Nash's proof is easy to read. What does he do? He works out the first Chern class of both bundles using the index theorem for families, and he gets something involving the Todd genus - and the Todd genus, as we all know, is defined using the same function
x / (1 - ex) = -1 + x/2 - x2/12 + ...
that we talked about in "week126" when computing the zero-point energy of the bosonic string! And yet again, it's that darn -1/12 in the power series expansion that makes everything tick. That's where the 13 comes from! It's all an elaborate conspiracy!
But of course the conspiracy is far grander than I've even begun to let on. If we keep digging away at it, we're eventually led to nothing other than....
MONSTROUS MOONSHINE!!!
But I don't have the energy to talk about that now. For more, try:
7) Richard E. Borcherds, What is moonshine?, talk given upon winning the Fields medal, preprint available as math.QA/9809110.
8) Peter Goddard, The work of R. E. Borcherds, preprint available as math.QA/9808136.
Okay, if you've actually read this far, you deserve a treat. First, try this cartoon, which you'll see is quite relevant:
9) Cartoon by J. F. Cartier, http://www.physik.uni-frankfurt.de/~jr/gif/cartoon/cart0785.gif
Second, let's calculate the determinant of an operator A whose eigenvalues are the numbers 1, 2, 3, .... You can think of this operator as the Hamiltonian for the wave equation on the circle, where we only keep the right-moving modes. As I already said, the zeta function of this operator is the Riemann zeta function. This function has zeta'(0) = -ln(2 pi)/2, so using our cute formula relating determinants and zeta functions, we get
det(A) = exp(-ζ'(0)) = (2 π)1/2 .
Just for laughs, if we pretend that the determinant of A is the product of its eigenvalues as in the finite-dimensional case, we get:
1 × 2 × 3 × ... = (2 π)1/2
or if you really want to ham it up,
∞! = (2 π)1/2.
Cute, eh? Dan Piponi told me this, as well as some of the other things I've been talking about. You can also find it in Bost's paper.
Récoltes et semailles
| 发表时间：2006-06-01, 01:18:51 |
作者资料
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kanex
发表文章数: 860
武功等级: 弹指神通
(第六重)
内力值: 343/343
Re: 无限维矩阵的行立式
actually, it's just zeta regularization. but it's really interesting.
Récoltes et semailles
| 发表时间：2006-06-01, 01:20:25 |
作者资料
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kanex
发表文章数: 860
武功等级: 弹指神通
(第六重)
内力值: 343/343
Re: 无限维矩阵的行立式
Notes and digressions:
In all of the above, I put a minus sign into my Laplacian, so that it has nonnegative eigenvalues. This is common among erudite mathematical physics types, who like "positive elliptic operators".
The zeta function trick for defining the determinant of the Laplacian works for any positive elliptic operator on a compact manifold. A huge amount has been written about this trick. It's all based on the fact that the zeta function of a positive elliptic operator analytically continues to s = 0. This fact was proved by Seeley:
10) R. T. Seeley, Complex powers of an elliptic operator, Proc. Symp. Pure Math. 10 (1967), 288-307.
Why is the Polyakov action <X, ΔX> conformally invariant? Because the Laplacian has dimensions of 1/length2, while the integral used to define the inner product has dimensions of length2, since the torus is 2-dimensional. This is the magic of 2 dimensions! The path integral for higher-dimensional "branes" has not yet been made precise, because this magic doesn't happen there.
Récoltes et semailles
| 发表时间：2006-06-01, 01:50:06 |
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那一剑的寂寞
发表文章数: 148
武功等级: 罗汉拳
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Re: 无限维矩阵的行立式
Kanex，你现在也在看Récoltes et semailles ？你看到哪里了？现在客栈很热闹，可是这本书比这热闹还有趣。我现在正在尝试着把它翻译成中文，可就是太多了，还只做了一点点。一起翻译，有兴趣否？
| 发表时间：2006-06-01, 10:05:33 |
作者资料
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kanex
发表文章数: 860
武功等级: 弹指神通
(第六重)
内力值: 343/343
Re: 无限维矩阵的行立式
我不懂法文，呵呵
英文网上有一点，不过直接看不是很好么
Récoltes et semailles
| 发表时间：2006-06-01, 12:23:01 |
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