User mkreisel - MathOverflow

The "right" $C^*$ algebraic proof of Bott Periodicity

2013-05-24T19:32:46Z

In learning about the K-theory of $C^*$-algebras, I have encountered the following 3 proofs of Bott periodicity:

$\bullet$ An argument based on Moyal quantization found in "Elements of Noncommutative Geometry."

$\bullet$ An argument based on Toeplitz algebras in Murphy's book.

$\bullet$ A seemingly brute force (but elementary) argument in Rordam, Larsen and Laustsen's book where they prove a variety of density results about projections in matrix algebras.

Which of these proofs is the most "geometrical" in the sense that it has a nice geometric interpretation when we restrict our attention to commutative $C^*$-algebras? As a student interested in noncommutative topology and geometry, if I wanted to study one of these proofs in gory detail, which should it be?

If there are other proofs that are more geometrical or more essential for understanding the phenomenon of Bott periodicity, I would be happy to hear about those too.

Meromorphic Functions as Distributions

2013-04-18T21:13:14Z

For the function $\frac{1}{x}$ on the real line, one can use a modified principal value integral to consider it as a distribution p.f.$(\frac{1}{x}),$ and one can do a similar construction to make $\frac{1}{x^m}$ into a distribution for $m>1.$ In the complex plane, the function $\frac{1}{z^m}$ is locally integrable for $m=1,$ but for larger $m$ some construction analogous to the one dimensional would have to be done to make it into a distribution.

More generally, given a meromorphic function on the plane (or torus), one should be able to consider it as a distribution by integrating against it and subtracting off some delta distributions or derivatives of delta distributions. Is this process explained in detail anywhere? Has anyone computed the Fourier series of such distributions, say for the Weierstrass $\mathfrak{p}$ function on the torus?

Linear Recurrence Relations in 2 Variables with Variable Coefficients

2013-04-05T23:34:24Z

Consider the following recurrence relation: $$-2a_{n,m} +a_{n-1,m}+a_{n,m-1}=0,$$ where $a_{n,m} \in \mathbb{C}.$I would like a purely combinatorial way to understand the subspace of solutions to this equation which have tempered growth. There is an obvious solution given by setting all $a_{n,m}=C$ for any constant $C,$ and I have reasons (coming from topology) to believe that these are the only solutions with tempered growth.

Now consider the similar but probably much harder recurrence relation: $$-2a_{n,m} +a_{n-1,m}+e^{2 \pi i n \theta}a_{n,m-1}=0$$ where $\theta$ is a fixed irrational. Note that the relation now depends on $n.$ I haven't even been able to come up with a solution to this that has tempered growth. I am hoping that it also has a 1 dimensional (or at least finite dimensional) space of solutions with tempered growth.

Is there a general combinatorial method for attacking either of these recurrence relations? Is there a general way to attack any linear recurrence relation like these?

EDIT: Let me also give my "proof" (I think it is correct) that any solution to the first relation with tempered growth must be constant. Consider the 2 dimensional torus thought of as $S^1\times S^1$, where $S^1$ is the unit circle. Now consider the function $z_1+z_2-2,$ thought of as a (finite) Fourier series. This has only 1 zero, at $(1,1).$

Now consider distributions $D$ on the torus, also thought of as Fourier series $D=\sum_\mathbb{Z^2} a_{n,m}z_1^nz_2^m$ where the $a_{n,m}$ now have only tempered growth. The first recurrence above is exactly the condition that $(z_1,+z_2 -2)D=0$ as a distribution. Since $z_1+z_2-2$ has only a single zero, the only solution with tempered growth should be a multiple of the Dirac distribution which is given by $a_{n,m}=1.$ I want a combinatorial proof or understanding of this phenomenon, since the second relation does not have this kind of topological interpretation. Ideally I would like to prove that the vector space of solutions to the second relation is also dimension 1, or is at least in some way related to the first relation.

2nd EDIT: WillSawin's answer shows that my initial proof is wrong. The space of tempered growth solutions to the first recurrence relation should be spanned (as a vector space) by the delta function and some linear combinations of its partial derivatives. Does the second recurrence have the same property? I.e. is there one "basic solution" $B$ to the second recurrence such that all other solutions can be expressed as linear combinations of the formal derivatives of $B?$

Moduli Spaces of Higher Dimensional Complex Tori

2013-04-03T00:15:34Z

I know that the space of all complex 1-tori (elliptic curves) is modeled by $SL(2, \mathbb{R})$ acting on the upper half plane. There are many explicit formulas for this action.

Similarly, I have been told that in the higher dimensional cases, the symplectic group $Sp(2n, \mathbb{R})$ acts on some such space to give the moduli space of complex structures on higher dimensional complex tori. Is there a reference that covers this case in detail and gives explicit formulas for the action?

In the 1-dimensional case, all complex tori can be realized as algebraic varieties, but this is not the case for higher dimensional complex tori. Does the action preserve complex structures that come from abelian varieties?

Crossposted at http://math.stackexchange.com/questions/345713/moduli-spaces-of-higher-dimensional-complex-tori where it has been unanswered for awhile.

Tiling Associated to an Almost Mathieu Operator

2013-03-04T16:57:41Z

In the references I've found, almost Mathieu operators of the form $$H_{\omega}^{\lambda, \alpha}u(n) = u(n+1)+u(n-1)+2\lambda\cos(2\pi(\omega+n\alpha))u(n)$$ acting on $l^2(\mathbb{Z})$ have been described as discrete aperiodic Schrodinger operators. Do they correspond in some way to aperiodic tilings of the real line? Physically this amounts to asking where the positions of the atoms are in this system.

Reference for Complex Abelian Varieties

2013-02-12T04:42:42Z

I am looking for a reference which explains how theta functions, algebraically independent meromorphic functions, and line bundles all fit together in the context of complex tori. More explicitly, given the right kind of Hermitian form on $\mathbb{C}^g$ with respect to some lattice $L$ I'd like an explicit construction of the g independent meromorphic functions on $\mathbb{C}^g/L$, which is something that Mumford, for example, does not give.

Answer by mkreisel for Differential geometry study materials

2013-02-08T22:58:45Z

I would recommend Lee's book "Introduction to Smooth Manifolds." It's a long book but is comprehensive, has complete proofs, and has lots of exercises.

Transforming the Dirac Operator on $S^1$

2012-12-13T18:35:26Z

This is related to my question http://math.stackexchange.com/questions/252742/transforming-the-dirac-operator-on-s1 on stack exchange which has not yet received an answer. For the purposes of this question, fix the spin structure over $S^1$ to be given by the connected double cover.

Sections of the bundle of spinors on $S^1$ can be thought of in two ways. First, by trivializing the bundle over $U_1 = S^1 \setminus {i}$ and $U_2 = S^1 \setminus {-i},$ sections can be considered as pairs of real valued functions $f_1, f_2$ defined on $\mathbb{R}$ such that $f_1(\frac{1}{x}) = f_2(x)$ when $x> 0,$ and $f_1(\frac{1}{x}) = -f_2(x)$ when $x< 0.$ Following Lawson/Michelson's "Spin Geometry," we can compute the Dirac operator locally. For example, $ f_1 \rightarrow i\frac{df_1}{dx}.$

We can also identify the sections as function $f: S^1 \rightarrow \mathbb{C}$ such that $f(-\theta) = -f(\theta)$ by the procedure described here http://math.stackexchange.com/questions/250835/expressing-the-sections-of-the-mobius-bundle-on-s1-as-globally-defined-real-v. When we identify sections this way, I have seen the Dirac operator expressed as $f \rightarrow -i\frac{df}{d\theta}.$ However I have never seen a derivation of this fact from the local expression given above. My attempt at the computation is contained in the stack exchange link at the top of the page.

Can anyone tell me how to translate between these 2 points of view?

What are some properties of Delone sets that come from Barlow packings of spheres?

2012-11-29T23:57:39Z

Given a Barlow packing of $\mathbb{R}^n$ by balls with at most a finite number of different radii, the centers of the balls will form a Delone set in $\mathbb{R}^n.$

For a highest density sphere packing, or at least a Barlow packing of highest density among Barlow packings, must the corresponding Delone set be a Meyer set? A Patterson set? In the cases of dimensions 2 and 3 where the optimal packing using a single radius is known, the sets can be chosen to be lattices.

I looked for papers exploring this connection and could only find this one https://www-fourier.ujf-grenoble.fr/PUBLIS/publications/REF_678.pdf, which is good, but doesn't address the big picture questions above.

EDIT: As the comments indicate, we should restrict to Barlow packings. In this case the Delone sets always appear to be of finite local complexity.

What are the Dirac operators on $S^1$?

2012-12-01T17:19:34Z

This is crossposted at stack exchange as http://math.stackexchange.com/questions/248391/dirac-operators-on-s1.

I am trying to understand the Dirac operators associated to the 2 spinor bundles on $S^1.$ I have been getting very confused about why one bundle has nontrivial harmonic spinors and the other doesn't.(Harmonic spinors are solutions $s$ to the equation $Ds = 0$ where $D$ is the Dirac operator and $s$ is a section.)

Here is my argument (which must be wrong somewhere). We have 2 spin structures given by the connected 2-fold covering and the disconnected 2-fold covering. Since the tangent bundle $TS^1$ is trivial, we can choose the trivial connection on it given by $f \rightarrow df.$ When considered as a connection on the principal bundle of frames (also isomorphic to $S^1$), i.e. as a Lie algebra valued one form on $S^1,$ it must be the zero form.

Ok, so now given either spin structure, the connection must lift to the $0$ connection. Furthermore, any complex line bundle over the circle is trivial, so both cases look exactly the same, and the Dirac operator appears to be $f \rightarrow i\frac{df}{dx}.$

However, I am told that in the case of the connected double cover we should have an additional condition on our $f,$ namely that it should satisfy $f(-x) = -f(x).$ With this extra condition, there cannot be harmonic spinors on the spinor bundle associated to the connected spin structure. Where have I gone wrong?

Explicit Computations of Examples in Spin Geometry

2012-12-12T01:26:48Z

I have been trying to learn about spin geometry, Dirac operators, and index theory by reading Lawson/Michelson's "Spin Geometry" and Friedrich's "Dirac Operators in Riemannian Geometry." Both are abstract, and basically no explicit examples are worked all the way through.

For example, I have been trying to find the spinor bundles, Dirac operators, and various indices for relatively simple manifolds: spheres and tori. However often these computations are detailed and even when I get to the end, it's not clear that I've done it correctly.

Is there another book, or perhaps online notes, which have a bunch of examples worked through in detail so that I can make sure what I'm doing is correct and also have a bank of examples to look at as I progress?

Spin structures on $S^1$ and Spin cobordism

2012-11-27T14:49:46Z

I'm trying to understand the 2 spin structures on the circle. Since the frame bundle for the circle is just the circle itself, Spin structures on $S^1$ correspond to double covers of $S^1$. There are two choices: the connected double cover and the disconnected double cover.

From the point of view of Spin cobordism, we can view the circle as the boundary of the disk in the plane. The disk has a unique spin structure, and we can ask which spin structure this induces on the boundary.

Lawson/Michelson's "Spin Geometry" claims that this induces the spin structure coming from the double cover, but I'm having trouble seeing that. The frame bundle for the disk $D^2$ must be trivial, and thus isomorphic to $D^2\times SO(2) = D^2 \times S^1.$ There is a natural double cover given again by $D^2 \times S^1,$ and the map is just the identity on $D^2$ and $z \rightarrow z^2$ on the $S^1$ factor.

To see what the induced spin structure on the boundary is, we must view the frame bundle of the boundary as sitting inside the frame bundle of $D^2\times S^1$ by fixing an outward normal vector field and then using it to complete any frame on $S^1$ to a frame on $D^2.$ To me, this seems to say that we view the frame bundle of $S^1$ (which is itself $S^1)$ as $S^1\times {1} \subset D^2 \times S^1,$ since once we fix one vector of a frame (in this case given by the normal) the other is entirely determined since we are in 2 dimensions.

But now if we look at the inverse image of that in the double cover, we appear to get two disjoint copies of $S^1,$ i.e. the disconnected double cover. What am I doing wrong?

(This is crossposted on stack exchange as http://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism).

Cut and Project Sets Using Hyperbolic Space

2012-11-19T17:54:53Z

One strategy for creating aperiodic sets in $\mathbf{R}$ is to take a line $L$ of irrational slope in $\mathbf{R}^2$ along with a compact window $W \subset \mathbf{R}$ which is thought of as a subset of $L^\perp.$ For simplicity, you can just make $W$ an interval containing 0. We then take any points of $\mathbf{Z}^2$ (say the standard lattice in ${\mathbf{R}^2}$) which are also contained in $L\times W$ and project them orthogonally onto $L.$ If we identify $L$ with $\mathbf{R}$ then we have constructed an aperiodic set in $\mathbf{R}$ which also has a variety of nice dynamical and analytic properties. This construction can be significantly generalized using abelian groups, but I was thinking of a slightly different generalization.

Why not do a similar construction using $\mathbf{H}^2?$ Namely, fix a uniform tiling of $\mathbf{H}^2$ and a geodesic $\gamma$ in $\mathbf{H}^2$ which descends to a geodesic on the (potentially singular) quotient surface. We would probably want to choose a geodesic which is not closed and is dense in some region of the surface. Choose some segment of a geodesic orthogonal to $\gamma$ to act like a window and project any vertices of the uniform tiling orthogonally onto $\gamma$ if they lie within the range of the window. (The region corresponding to $L \times W$ in the hyperbolic setting would look something like a banana in the Poincare disc model of $\mathbf{H}^2$.) After composing with an isometry from $\gamma$ to $\mathbf{R}$ we might get an aperiodic point set.

Is there an obvious reason why this wouldn't yield (at the very least) an aperiodic tiling of $\mathbf{R}$ with finite local complexity?

Comment by mkreisel

2013-05-24T20:28:15Z

Certainly the arguments there would be the most geometrical. I guess I was hoping to see how "geometry" was present in a proof that worked for all $C^*$-algebras and not just topological spaces.

Comment by mkreisel

2013-04-19T14:25:56Z

For a doubly periodic meromorphic function (so a function on the torus) does this gluing procedure ensure that the resulting distribution will also be doubly periodic? It seems like it should, since in any neighborhood not containing a pole it is given by integration against a doubly periodic function.

Comment by mkreisel

2013-04-08T18:19:11Z

Ahhh I see. I will think some more about it. Thanks again for the help!

Comment by mkreisel

2013-04-08T17:28:31Z

Furthermore, if we choose the initial sequence $b_n=a_{n,-n}$ to be bounded, then clearly $|T^kb_n|=|a_{n,k-n}|$ will be bounded in norm by the same bound. Were you proving that every sequence has exponential growth? If so it couldn't have been coming from applying $T,$ as the above counterexample shows, and I don't see how your analysis can take into account the lower left portion, as $T$ doesn't "see" that. If you were only proving that some (maybe many?) solutions have exponential growth then I agree with you, but did those solutions already have initial data with exponential growth?

Comment by mkreisel

2013-04-08T17:22:28Z

Sorry if I'm being obtuse, but I'm still having trouble understanding your solution. I now understand how the operator $T$ helps; if $a_{n,m}$ satisfies the recurrence relation I can take $b_n=a_{n,-n}$ to be my "initial data" and then applying $T$ yields the values $Tb_n=a_{n,1-n}.$ So understanding the dynamics of $T$ tells me about the growth of the sequence as we move up and to the right in the $\mathbb{Z}^2$ lattice. However it cannot tell us about the lower left portion of the grid...

Comment by mkreisel

2013-04-07T14:52:31Z

How does your first step work? The original recurrence gives $$a_{n,m}=\frac{a_{n-1,m}+a_{n,m-1}e^{2\pi i n\theta}}{2}$$ but you've written $$a_{n,m}=\frac{a_{n-1,m-1}+a_{n,m-1}e^{2\pi i n\theta}}{2}.$$

Comment by mkreisel

2013-04-06T21:28:23Z

Thanks! These solutions correspond to the fact that the partial derivatives agree at $(1,1).$

Comment by mkreisel

2013-04-06T19:37:48Z

As I noted in the comment above, specifying the $S_0$ plus the values of $a(0,n), n<0$ will completely determine the grid for either relation. Clearly if we make either of these choices in a way that does not have tempered growth then the solution they determine will not have tempered growth. I think the solutions you describe above were essentially chosen, as I described, not to have tempered growth. What is more interesting to me is whether choosing bounded or temperate growth initial data leads to a bounded/temperate growth solution.

Comment by mkreisel

2013-04-06T19:06:09Z

However, in this case we have $a_{-n,0} = 2^n,$ so this solution clearly cannot be tempered since it is growing exponentially in $n$ in this direction. To see that $a_{-n,0}=2^n,$ note that because we have placed zeros on the half-column $a_{0,-m},$ the strict lower left quadrant $m,n<0$ will be entirely zero, and in particular the half-row $a_{m,-1} = 0$ for $m<1.$ Thus since we placed a 1 at $a_{0,0},$ we get $a_{-1,0}=2,$ and since the entire half-row is zero we get $a_{-n,0} = 2^n.$ So just specifying a bounded initial data does not mean the whole sequence remains bounded, nor tempered.

Comment by mkreisel

2013-04-06T18:57:18Z

I do not think this is quite right. I agree that if you specify $S=\{(-i,i)\}_{i \in \mathbb{Z}}$ then the region above and to the right of $S$ is determined and bounded. However your extension to the lower left does not make sense to me. It is not clear to me how specifying $T=\{(i,i)_{i<0}\}$ determines the rest of the grid. So instead consider this example. Let $a_{n,m} = 1$ on $S$, and thus $a_{n,m}=1$ whenever $n+m>1$ as well. Now let $a_{0,m}=0$ for all $m<0,$ similar to what you have described. This fully determines the grid as we can now work down the lower diagonals one at a time...

Comment by mkreisel

2013-02-23T14:51:49Z

Is this condition sufficient? Are there "fake groups" or "fake H-spaces" with the right structure on cohomology but no actual group structure?

Comment by mkreisel

2012-12-23T17:28:04Z

Even a comment outlining how you would do the calculation, even if it doesn't show the calculation in full, would be helpful.

Comment by mkreisel

2012-12-20T15:21:23Z

Thanks! It seems like this question hasn't yet been fully explored...

Comment by mkreisel

2012-12-15T03:58:09Z

Thanks, that helps a lot!

Comment by mkreisel

2012-12-15T00:37:32Z

You're right, changed it.