User liviu nicolaescu - MathOverflow

Answer by Liviu Nicolaescu for Karoubi versus Kasparov K-theory

2013-05-21T10:04:41Z

About 15 years ago I used Karoubi's description of $K$-theory to solve prove some cutting and pasting formula for the index families of elliptic problems. To do so I needed rephrase Karoubi's theory into something more flexible and more computable. Some of the interpretations I found might be relevant to your question. I will briefly describe one such interpretation referring for proofs and many more details to the original source, my old paper Generalized symplectic geometries and the index of families of elliptic problems, Mem. A.M.S., vol. 128, no.609, 1997. I will refer to this as the old paper. $\newcommand{\bsH}{\mathscr{H}}$

To simplify the presentation let me define a $C^{p,q}$-module to be a Hilbert space $H$ equipped with a $8$_morphism of $C^*$_algebras $\phi: C^{P,q}\to B(H)$. A graded $C^{p,q}$-module can then be identified witha $C^{p,q+1}$-module. $\newcommand{\eF}{\mathscr{F}}$

Suppose that $H$ is a $C^{p,q+1}$-module. Define $\eF^{p,q}$ (or $\eF^{p,q}(H)$ to be the space of closed, densely defined, Fredholm, selfadjoint operators $T: H\to H$ that super-commute with the $C^{p,q}$-structure, i.e.,

$$ Te_k+ e_k T=0,\;\;\forall k=1,\dotsc, p+q. $$

The space $\eF^{p,q}$ carries a natural topology defined by the metric

$$ d(T_1,T_2)= \Vert \Psi(T_1-\Psi(T_2)\Vert,\;\;\Psi(\lambda=\lambda(1+\lambda^2)^{\frac{1}{2}}. $$

Denote by $\newcommand{\eBF}{\mathscr{BF}}$ $\eBF^{p,q}$ the subspace of $\eF^{p,q}$ consisting of bounded operators. Then one can show (see here) that the inclusion $\eBF^{p,q}\hookrightarrow \eF^{p,q}$ is a homotopy equivalence and that $\eBF^{p,q}$ is a classifying space for Karoubi's $KO^{p,q}$, which for simplicity I will denote by $K^{p,q}$.

Thus, to a compact $CW$-complex and a continuous map $T: X\to\eF^{p,q}$ one can associate an element $(E,\eta_0,\eta_1)\in K^{p,q}(X)$. To effectively describe this correspondence

$$ (X\stackrel{T}{\to}\eF^{p,q}) \to (E,\eta_0,\eta_1), $$

one needs a new, symplectic description of Karoubi's $K$-theory, and it is through the symplectic prism that I got to see Kasparov's KK lurking in the background.

Let $T\in \eF^{p,q}(H)$. Recall that, by construction $H$ is a $C^{p,q+1}$-mpodule. The direct sum $\hat{H}=H\oplus H $ has a richer structure of $C^{p+1,q+1}$-module (see section 5.2 in the old paper)

The graph $\Gamma_T$ of $T$ is a closed subspace of $\hat{H}$. Denote by $R_T:\hat{H}\to\hat{H}$ the orthogonal reflection in $\Gamma_T$. Observe that the subspace $H\oplus 0\in \hat{H}$ can be identified with $\Gamma_0$, the graph of the trivial linear map. We set

$$\Gamma_\infty=0\oplus H\subset \hat{H}. $$

Then $R_T^2=1$, and the condition $T\in\eF^{p,q}$ is equivalent with the following requirements.

$R_T$ supercommutes tith the $C^{p+1,q+1}$-structure on $\hat{H}$.
The pair of subspaces $(\Gamma_0,\Gamma_T)$ is a Fredholm pair.
The subspace $\Gamma_T$ does not intersect $\Gamma_\infty$.

$\newcommand{\eFL}{\mathscr{FL}}$. We denote by $\eFL^{p+1,q+1}(\hat{H})$ the set of closed subspaces $L$ of $\hat{H}$ such that the reflection $R_L$ in $L$ satisfies the conditions $1$ and $2$ above. (In the old paper I called these spaces generalized lagrangian spaces of type $(p+1,q+1)$. We have thus produced an inclusion

$$\eF^{p,q}(H)\to \eFL^{p+1,q+1}(\hat{H}). $$

One can show two things. First, the space $\eFL^{p+1,q+1}$ classifies $K^{p+1,q+1}$ and second, the above inclusion is a homotopy equivalence. (The proof uses Bott periodicity.) One can use a process of symplectic reduction to canonically associate to continuous family $L: X\to \eFL^{p+1,q+1}(X)$ an element

$$ (E,\eta_0,\eta_1)\in K^{p+1,q+1}(X)\cong K^{p,q}(X). $$

Observe that the elements of $\eFL^{p+1,q+1}$ can be identified with selfajoint operators $R:\hat{H}\to \hat{H}$ such that $R^2=1$ and super-commute with the $C^{p+1,q+1}$-structure and they satisfy condition 2. This almost looks like a Kasparov element.

To actually get a Kasparov element consider a smooth, odd, nondecreasing function $\newcommand{\bR}{\mathbb{R}}$

$$\beta : \bR\to \bR,\;\;\beta (t)=\pm 1\;\;\mbox{if $\pm t>1$}. $$

For $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ we set $\beta_\ve(t)=\beta(t/\ve)$.

Then for any $T\in\eF^{p,q}$, $\beta_\ve (F)$ defines a Kasparov element for $\ve>0$ sufficiently small.

Remark. I will describe below an explicit map

$$K^{p,q}(X)\to [X, \eF^{p+1,q+1}]. $$

This is more or less what you need since $K^{p,q}(X)\cong K^{p+1,q+1}(x)$ and more generally $K^{p_0,q_0}(X)\cong K^{p_1,q_1}(X)$ if $p_0-q_0\equiv p_1-q_1\bmod 8$.

The symplectic point of view is crucial since the above map is inspired by Floer's work on symplectic (Floer) homology. To justify the symplectic terminology let me discuss a simple example.

Let us look at a $C^{1,0}$-module. This is a real Hilbert space equipped with a orthogonal operator $J: H\to H$ such that $J^2=-1$, $J^*=-J$. In the finite dimensional case think $H=\bR^n\oplus \bR^n$, $J(x,y)=-(y,x)$.) A subspace $L\subset H$ is called Lagrangian if $JL=L^\perp$.

In the case of $\bR^n\oplus \bR^n= T^*\bR^n$ equipped with the canonical symplectic structure we obtain in this fashion the classical notion of Lagrangian subspace. If $P_L$ denotes the orthogonal projection onto $L$ and $R_L=2P_L-1$ denotes the orthogonal reflection in the subspace $L$, then $L$ is Lagragian iff $R_L$ anticommutes with $J$,

$$R_LJ+JR_L=0. $$

In algebraic terms $R_L$ defines a $\newcommand{\bZ}{\mathbb{Z}}$ $\bZ/2$-grading of the $C^{1,0}$-module $H$. However the symplectic point of view has more flexibility because it leads to certain operations which algebraically do not seem natural. (I'm thinking here of the process symplectic reduction.)

In general given a $C^{p,q}$-module $H$, we define a $(p,q)$-Lagrangian in $H$ to be a subspace $L\subset H$ such that $R_L$ supercommutes with the $C^{p,q}$-structure, i.e., $R_L$ is a $\bZ/2$-grading of the $C^{p,q}$-module $H$. Denote by $\DeclareMathOperator{\Lag}{Lag}$ $\Lag^{p,q}(H)$ the space of $(p,q)$-lagragians in $H$. Observe that and element in $K^{p,q}(X)$ is defined by a continuous map

$$ X\to {\Lag}^{p,q}(H)\times {\Lag}^{p,q}(H), \;\;x\mapsto (L_0(x), L_1(x))$$

where $H$ is a finite dimensional $C^{p,q}$-module. Thus an element in $K^{p,q}$ is a pair of continuous families of lagrangian subspaces in a $C^{p,q}$-module. Moreover, it suffices consider only the case when one of the families $L_0(x)$ is constant.

Fix a finite dimensional $C^{p',q'}$-module $H$, $p'=p+1$, $q'=q+1$. Denote by $J$ the operator on $H$ defined by the multiplication by $e_{p+1}$ so that $J^*=-J$, $J^2 =-1$.

To a pair of lagrangians $L_0, L_1\in \Lag^{p',q'}(H)$ we associate an operator $T_{L_0,L_1}\in\eF^{p,q}(\bsH)$ where

$$\bsH=L^2(0,1, H), $$

and $T_{L_0,L_1}$ is the closed, Fredholm selfadjoint unbounded operator on $\bsH$ with domain

$$ D(T_{L_0,L_1})=\bigl\lbrace u\in L^{1,2}(0,1; H);\; u(0)\in L_0,\;\;u(1)\in L_1\bigr\rbrace, $$

such that

$$ T_{L_0,L_1} u(s)=J\frac{du}{ds},\;\;s\in (0,1). $$

Above, $L^{1,2}$ denotes the Sobolev spaces of functions with first order derivative in $L^2$.

The motivation for the operator $T_{L_0,L_1}$ comes from symplectic Floer homology. In his papers on the (Floer) homology of a pair of lagrangian submanifolds A. Floer investigated the operators $T_{L_0,L_1}$ in the case $p=1, q=0$ and the indices of one-parameter families of such operators. Note that only the domain of $T_{L_0,L_1}$ depends on $L_0,L_1$. The action of $_{L_0,L_1}$ is independent of the lagrangians $L_0,L_1$>

To an element $\alpha\in K^{p',q'}(X)$ represented by a continuous map

$$ X\ni x \to (L_0(x), L_1(x))\in {\Lag}^{p',q'}(H)\times {\Lag}^{p',q'}(H) $$

we can associate an element $T_\alpha\in [X,\eF^{p,q}]$ given by the continuous map

$$ X\ni x\mapsto T_{L_0(x), L_1(x)}\in\eF^{p,q}(\bsH). $$

As mentioned before, $\eF^{p,q}(\bsH)$ classifies $K^{p,q}$ and thus the map $T_\alpha$ defines an element ${\rm ind}\; T_\alpha\in K^{p,q}(X)$. In that old paper I proved that ${\rm ind}\; T_\alpha \in K^{p,q}(X)$ coincides with $\alpha\in K^{p',q'}(X)$ via the canonical isomorphism $K^{p,q}(X)\to K^{p',q'}(X)$. The proof uses crucially the process of symplectic reduction. For details see Thm. 5.5 in the old paper.

The familly $ T_{\alpha}$ can be given a Kasparov descriptition as explained above.

Answer by Liviu Nicolaescu for objects which can't be defined without making choices but which end up independent of the choice

2013-05-20T21:22:31Z

The Seiberg-Witten invariant of an oriented compact manifold $4$-manifold needs at least a metric and often an additional $2$-form to be defined. Ultimately it is independent of the choice of metric and form

Answer by Liviu Nicolaescu for Circle bundles over $CP^1$ and self-intersection number of $CP^1$ embeddings

2013-05-15T20:22:48Z

Take your favorite complex line bundle $L$ over $\newcommand{\bCP}{\mathbb{CP}}$ $\bCP^1$. For the rank two complex vector bundle $E\to\bCP^1$ defined as the direct sum of $L$ with the trivial line bundle $\newcommand{\uc}{\underline{\mathbb{C}}}$ $\uc\to\bCP^1$

$$ E=L\oplus \uc. $$

Now form the projectivization $\newcommand{\bP}{\mathbb{P}}$ $\bP(E)$. This is a bundle over $\bCP^1$ whose fiver over a point $x\in\bCP^1$ is the projective space $\bP(E_x)$ consiting of all complex lines in the vector space $E_x=L_x\oplus \mathbb{C}$.

The $0$-section of $L$ defines a section $\zeta: \bCP^1\to\bP(E)$. The normal bundle of the embedding $\zeta(\bCP^1)\subset \bP(E)$ can be identified with the line bundle $L$.

Continuous dependence of the expectation of a r.v. on the probability measure

2013-05-13T15:38:59Z

$\newcommand{\bsV}{\boldsymbol{V}}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\bR}{\mathbb{R}}$ Suppose that $\bsV$ is an $N$-dimensional real Euclidean space. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the space of symmetric positive semidefinite operators $A:\bsV\to \bsV$. To each $A\in \eA$ we can associate in a canonical fashion a centered Gaussian measure $\gamma_A$ on $\bsV$ which is concentrated on $(\ker A)^\perp$. For example, if $A$ is nondegenerate, then

$$ \gamma_A(dv)= \frac{1}{\sqrt{\det 2\pi A}} e^{-\frac{1}{2} (A^{-1}v,v)}dv, $$

while if $A=0$, then $\gamma_0$ is the Dirac measure concentrated at the origin.

Fix a locally Lipschitz function $f:\bsV\to\bR$ which is positively homogeneous of degree $\alpha \geq 2$. For any $A\in \eA$ we denote by $\bsE_A(f)$ the expectation of $f$ with respect to the probability measure $\gamma_A$ on $\bsV$. Consider the function

$$ \eA\ni A\mapsto \bsE_A(f)\in \bR. $$

This function is continuous and positively homogeneous of degree $\frac{\alpha}{2}$, i.e.,

$$ \bsE_{tA}(f)=t^{\frac{\alpha}{2}} \bsE_A(f),\;\;\forall t>0,\;\;A\in\eA. $$

I am interested in its modulus of uniform continuity on the ball

$$\eA_1:=\bigl\lbrace A\in\eA;\;\;\Vert A\Vert\leq 1\bigr\rbrace, $$

I was able to prove that on this ball the above function is Holder continuous, with Holder exponent $\frac{1}{2N+3}$. This suffices for the applications I have in mind, but I strongly suspect that it is far from optimal. I believe that the Holder exponent $\frac{1}{2}$ is uniformly optimal in the following sense: there exist $C, r>0$ so that for any $A, B\in\eA_1$ satisfying

$$\Vert A- B\Vert \leq r, $$

we have

$$\bigl\vert \bsE_A(f)-\bsE_B(f)\bigr\vert\leq C\Vert A-B\Vert^{\frac{1}{2}}. \tag{1} $$

Remark. To see that the exponent $\frac{1}{2}$ is the best one can hope for consider the case $\bsV=\bR^2$, $f(x,y)=|xy|$ and $\newcommand{\ve}{{\varepsilon}}$ and the Gaussian measures

$$ \gamma_{A_\ve}=\frac{1}{2\pi\ve} e^{-\frac{1}{2\ve^2}x^2-\frac{1}{2}y^2} |dxdy| $$

Then $\Vert A_\ve-A_0\Vert =\ve^2$,

$$\bsE_{A_0}(f)=0,\;\; \bsE_{A_\ve}(f)=\left(\int_{\bR}|x|e^{-\frac{1}{2}x^2} |dx|\right)^2 \ve. $$

My question is the following: have you encountered continuity results of this sort, and if so, can you indicate some references that deal with this? Thanks!

Answer by Liviu Nicolaescu for Trivial Line Bundle-Riemann surfaces

2013-05-10T12:52:21Z

The trivial line bundle $\newcommand{\bC}{\mathbb{C}}$ $\underline{\bC}_M:=\bC\times M\to M$ over a complex manifold $M$ admits a trivial metric $h_0$. This is defined by requiring that the trivial section $u_0$

$$ M\ni p\mapsto u_0(p)=1\in\bC$$

has pointwise length $1$. If $h$ is another metric on $\underline{\bC}_M$, then $h= w^2\cdot h_0$ where $w$ is the positive function $w(p)=|u_0(p)|_h$, $p\in M$. The function $w$ thus can be expressed as an exponential $w= e^{-\varphi/2}$,

$$\varphi=\log |u_0|_h^2. $$

As shown in many books on complex differential geometry (e.g. Griffiths and Harris) the curvature of this line bundle is

$$ \bar{\partial}\partial \log |u_0|_h^2= - \bar{\partial}\partial \varphi. $$

Answer by Liviu Nicolaescu for Enumerating/counting paths of a given length on a 2D lattice

2013-05-06T22:17:59Z

I am trying to make sense of your problem. From what I can gather you are talking about walks with steps of size $1$ in each of the cardinal directions East, West, North, South, (E,W,N,S). To use your notation $E=x$, $W=X$, $N=y$, $S=Y$. You are not allowed to immediately backtrack, i.e., if say you take an $E$-step, then your next step cannot be a $W$-step. I will refer to such paths as admissible. I think that you need to fix the starting point. Assume it is the origin.

The number of admissible paths of length $n$ starting at the origin is $4\cdot 3^{n-1}$.

Alas, there is a symmetry in the problem and this is where I am a bit confused. Its looks to me that the symmetry group is the group $G$ of symmetries of the square with vertices $(\pm 1,0)$, $(0,\pm 1)$. (I could be wrong, but that is what I am getting from your description.) The center of this square is the origin, an the vectors obtained by joining the center with the vertices are the unit vectors $\vec{E},\vec{W},\vec{N},\vec{S}$ pointing in the four cardinal directions.

The group has eight elements and it is generated by the reflection $R_x$ in the $x$-axis and the counterclockwise rotation $J$ by ninety degrees. The elements of this group are

$$ 1, J, J^2, J^3, R_x, R_xJ, R_xJ^2, R_xJ^3. $$

Assuming that my guess is correct you need to count admissible paths, where two paths that are related by one of the above eight symmetries are considered identical. Denote by $Q_n$ the number of such paths of length $n$.

Here you need to invoke Burnside's theorem. Here is what it says in this case.

For each $g\in G$ denote by $P_n(g)$ the number of admissible paths of length $n$ that admit $g$ as symmetry. More precisely a path

$$v_1\dotsc,v_n, \;\;v_1,\dotsc,v_n\in \lbrace E,W,N,S\rbrace $$

admits $g$ as symmetry if $g(v_1)\dotsc g(v_n)=v_1\dotsc v_n$. Then Burnside's theorem states that

$$Q_n= \frac{1}{|G|}\sum_{g\in G} P_n(g). $$

There are only three elements $g\in G$ for which $P_n(g)\neq 0$. They are $1, R_x, R_y$, where $R_y$ denotes the reflection in the $y$-axis. Putting all the above together we deduce that

$$Q_n= \frac{1}{8}\Bigl( 4\cdot 3^{n-1}+ 2+2)=\frac{1}{2}\bigl(3^{n-1}+1\bigr). $$

Update. I have worked out the details taking into account the correct symmetry group.

Here is what I found. If $n$ is odd, $n=2m-1$, then

$$ Q_n=\frac{1}{4}\bigl(3^{n-1}+2\cdot 3^{m-1}+1\bigr). $$

If $n$ is even, $n=2m$, then

$$ Q_n = \frac{1}{4}\bigl( 3^{n-1}+4\cdot 3^{m-1}+1\bigr). $$

Here are a few values.

$$Q_1=1, \;\; Q_2=2,\;\; Q_3=4, \;\; Q_4= 10,\;\; Q_5=25,\;\; Q_6=70. $$

Note that I get $Q_4=10\neq 8,9$. In any case, the details can be found here. Maybe somebody can explain the discrepancy involving $Q_4$.

Answer by Liviu Nicolaescu for Modern Mathematical Achievements Accessible to Undergraduates

2013-05-05T20:14:28Z

Google PageRank algorithm is remarkable mathematics we experience daily. It's even taught in some undergraduate classes. Here is the link to one such class

http://www.math.harvard.edu/~knill/teaching/math19b_2011/index.html

Answer by Liviu Nicolaescu for How to compute the Alexander polynomial of general torus knot

2013-05-05T14:44:54Z

I think that the computation of the Alexander polynomial of torus knots and more general algebraic knots goes back to Bureau. There is a general trick called the Seifert-Torres formula that allows you to compute the desired Alexander polynomial of the $(p,q)$-torus knot and much more. For a particularly an elegant proof based on the concept of Reidemeister torsion I refer to Turaev's most excellent survey Reidemeister torsion in knot theory, Russian Math. Surveys, vol. 41 (1986), 119-182. The concept of Reidemeister torsion is what hides behind the Alexander polynomial so it's worth having a look at this concept.

Answer by Liviu Nicolaescu for Does a topological manifold have an exhaustion by compact submanifolds with boundary?

2013-05-01T17:10:40Z

Whitney embedding theorem shows that any connected smooth manifold $M$, compact or not, admits a proper imbedding $\newcommand{\bR}{\mathbb{R}}$ into an Euclidean space $\bR^N$ where properness signifies that the intersection of the image of the embedding with any compact set is a compact set.

Assume that $M\subset \bR^N$ is properly embedded. For a point $q\in \bR^N$ define $f_q:M\to\bR$ by setting

$$ f_q(p)= |p-q|^2,\;\;\forall p\in M. $$

Since $M$ is properly embedded we deduce that sublevel sets $\lbrace f_q\leq c\rbrace\subset M$ are compact for any $t\in\bR$.

For generic $q\in\bR^N$ the function $f_q: M\to\bR$ is Morse. Fix such a $q$. Thus each sublevel set $\lbrace f_q\leq t\rbrace$ contains finitely many critical points. This implies that the set of critical values of $f$ is a discrete countable subset of $\bR$.

Choose an increasing and unbounded sequence $(r_n)_{n\geq 1}$ of regular values of $f_q$ and set

$$M_n:=\lbrace f_q\leq r_n\rbrace. $$

The collection $(M_n)_{n\geq 1}$ is an exhaustion of $M$ by compact manifolds with boundary.

On a differential inequality

2013-04-30T19:38:03Z

The question has probabilistic origins, but it would take too long to elaborate. $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$

Fix a nonnegative Schwartz function $w:\bR\to \bR$. For any positive integer $m$ let $V_m:\bR^m\to\bR$ denote the Fourier transform of

$$ w_m:\bR^m\to\bR,\;\; w_m(x)=w(|x|^2/2). $$

More precisely

$$ V_m(\xi)=\int_{\bR^m} e^{-\ii (x,\xi)} w(|x|^2/2) dx. $$

$V_m$ is an $O(m)$-invariant Schwartz function on $\bR^m$ so that it has the form

$$ V_m(\xi)= f_m\bigl(|\xi|^2/2\bigr), $$

where $f_m$ is a smooth one-variable function, $[0,\infty)\ni r\mapsto f_m(r)\in\bR$.

The function $f_m=f_{m,w}$ depends on the initial $w$. The dependence $w\mapsto f_{m,w}$ is linear and can be explicitly described in terms of the Hankel transforms.

Denote $\newcommand{\eF}{\mathscr{F}}$ by $\eF$ the class of $C^2$-functions $f:[0,\infty)\to \bR$ such that

$$ f'(0)< f'(r)+2rf''(r)<-f'(0),\;\;\forall r>0. \tag{1} $$

If we set $r=t^2/2$, $g(t)= f(r)=f(t^2/2)$, then $$ g''(t)=f'(t^2/2)+t^2f''(t^2/2)=f'(r)+2rf''(r), $$

and we can rephrase the above inequality as

$$ |g''(t)|<|g''(0)|,\;\;\forall t>0. \tag{2}$$

Denote by $\newcommand{\eW}{\mathscr{W}}$ $\eW$ the collection of nonnegative Schwartz functions $w :\bR\to \bR$ such that

$$f_{m,w}\in\eF, \;\;\forall m>0. $$

The description (2) and the $O(m)$ invariance of $V_m(\xi)$ lead to the following equivalent description of $\eW$. More precisely $w\in\eW$ iff for any $m>0$

$$|\Delta V_m(\xi)|<|\Delta V_m(0)|,\;\;\forall \xi\in\bR^m\setminus 0, \tag{3}$$

i.e.,

$$ \left\vert\int_{\bR^m} e^{-\ii(x,\xi)}|x|^2w\Biggl(\frac{|x|^2}{2}\Biggr) |dx|\right\vert <\int_{\bR^m} |x|^2w\Biggl(\frac{|x|^2}{2}\Biggr)|dx|,\;\;\forall\xi\in\bR^m\setminus 0. $$

Remark 1. The class $\eF$ contains all the completely monotone functions $f:[0,\infty)\to \bR$ such that $f''(0)>0$. (Recall that a function $f:[0,\infty)\to \bR$ is completely monotone if it is smooth and $(-1)^kf^{(k)}(t)\geq 0$, $\forall t\geq 0$, $\forall k\geq 0$.)

This follows from the following observations.

The functions $r\mapsto g_s(r)= e^{-sr}$, belong to $\eF$ for any $s>0$.
The set $\eF$ is a convex cone.
By Bernstein theorem, any completely monotone function $f$ can be written as an infinite superposition of nonnegative multiples of the functions $g_s$. More precisely, there exists a finite positive Borel measure $\mu(|ds|)$ on $[0,\infty)$ such that

$$ f(r)=\int_0^\infty e^{-sr} \mu(|ds|). $$

Remark 2. If $w:\bR\to\bR$ is a nonnegative Schwartz function whose restriction to $[0,\infty)$ is completely monotone, then $w\in \eW$.

Indeed, we can find a positive, finite Borel measure $\mu(|ds|)$ on $[0,\infty)$ such that

$$ w(t)= \int_0^\infty e^{-st}\mu(|ds|). $$

Then

$$ w_m(x)= \int_0^\infty e^{-s|x|^2/2}\mu(|ds|), $$

$$ V_m(\xi)= \int_0^\infty\left(\int_{\bR^m} e^{-\ii(x,\xi)} e^{-s|x|^2/2} dx\right) \mu(|ds|) $$

$$ = \int_0^\infty\left(\int_{\bR^m} e^{-\ii(y,\xi)/\sqrt{s}} e^{-|y|^2/2} dy\right) s^{-\frac{m}{2}}\mu(|ds|) = (2\pi)^{\frac{m}{2}}\int_0^\infty e^{-\frac{|\xi|^2}{2s}} s^{-\frac{m}{2}}\mu(|ds|). $$

Hence

$$ f_m(r)= (2\pi)^{\frac{m}{2}}\int_0^\infty e^{-\frac{r}{s}} s^{-\frac{m}{2}}\mu(|ds|).$$

This proves that $f_m(r)$ is completely monotone. A simple computation shows $f_m''(0)>0$ and from Remark 1 we conclude $f_m\in\eF$.

Now comes the question.

Does the class $\eW$ contain examples of functions $w$ not covered by Remark 2?

Remark 3. I should perhaps mention here a theorem of Schoenberg which states that a function $w:[0,\infty)\to \bR$ is completely monotone if and only if, for any $m>0$ the function

$$ \bR^m\ni x\mapsto w_n(x)=w(|x|^2/2) \in\bR$$

is positive definite. According to Bochner's theorem, this means that $w_m$ is the Fourier transform of a positive measure on $\bR^m$. In view of the above discussion we see we can rephrase the question as follows.

Is it true that $w\in \eW \Rightarrow f_{m,w}\geq 0$, $\forall m>0$?

Answer by Liviu Nicolaescu for Matrices whose kernel escapes a sub-vector space

2013-04-30T12:37:46Z

Denote by $\newcommand{\bC}{\mathbb{C}}$ $L(\bC^m\to \bC^n)$ the space of linear operator $\bC^m\to\bC^n$. Then $X$ is the complement in $L(\bC^m\to\bC^n)$ of the vector subspace

$$ \bigl\lbrace A\in L(\bC^m\to\bC^n);\;\; Av=0,\;\;\forall v\in V\bigr\rbrace\subset L(\bC^m,\bC^n). $$

In particular, $X$ is open in $L(\bC^m,\bC^n)$ (as the complement of a closed set) and constructible.

Answer by Liviu Nicolaescu for Good book on Calculus of Variations

2013-04-29T17:54:41Z

The book of Gelfand and Fomin is a good place to start. It worked for me. I would like to include another nice and short source namely Chapter 19, vol. II of Feynman's Lectures on Physics.

If you know a little about smooth manifolds, then Arnolds's Mathematical Methods of Classical Mechanics is another excellent source. Also, check volume in of Dubrovin, Fomenko, Bovikov, Modern Geometry.

Answer by Liviu Nicolaescu for Volume of Gr(2,4)

2013-04-27T15:12:35Z

Check section 9.1.2 of these notes There I compute the volumes of real Grassmannians. A similar computation works in the complex case.

Update Using the description $\mathrm{Gr}\;(k, N)\cong U(N/U(k)\times U(N-k)$ and a bi-invariant metric on $U(N)$, this induces bi-invaraint metrics on $U(k),U(N-k)\subset U(n)$ and an invariant metric on $\mathrm{Gr}(k,N)$. The volume of $\mathrm{Gr}(k,N)$ with respect to this metric is

$$ {\rm vol} \mathrm{Gr}(k, N)= \frac{ {\rm vol}\; U(N)}{{\rm vol}\; U(k)\cdot {\rm vol}\; U(N-k)}. $$

The volume of a compact Lie group $G$ with respect to a bi-invariant metric $g$ was computed by I.G. Macdonald,

The volume of a compact Lie group, Invent. Math. 56(1980), no. 2, 93–95.

For the Lie group $U(n)$ this takes the form

$$ {\rm vol}\; U(n)=\frac{1}{(2P_n)^2(2\pi)^n}\times {\rm vol}\; T^n\times \prod_{k=1}^n {\rm vol}\;S^{2k-1}, $$

where ${\rm vol}\; T^n$ denotes the volume of the maximal torus of $U(n)$ equipped with the induced bi-invariant metric, and $P_n$ is the product of the lengths of the positive roots of $U(n)$.

Answer by Liviu Nicolaescu for Visualize Fourth Homotopy Group of $S^2$

2013-04-26T14:19:38Z

For me, the most satisfactory explanation of the isomorphism $\pi_4(S^2)\cong \mathbb{Z}/2$ is in Pontryagin's beautiful survey

Smooth manifolds and their applications in homotopy theory. 1959 American Mathematical Society Translations, Ser. 2, Vol. 11 pp. 1–114 American Mathematical Society, Providence, R.I.

He derives the isomorphism by using the Pontryagin-Thom construction. Pontryagin has a soft touch and patience with the details. In Section 15 he deals with $\pi_{n+2}(S^n)$. I had a very enjoyable intellectual experience reading this paper.

Answer by Liviu Nicolaescu for Support of a module over a polynomial algebra

2013-04-25T10:05:59Z

This is more or less a classical result but I believe some comments are warranted. Atiyah and Bott are first and foremost geometers/topologists and they opted for a more geometric description of the concept of support better adapted to the purposes of that (wonderful) paper.

One can define the support of a module over an arbitrary ring $R$. It is a subset of $\mathrm{spec} (R) =$ the set of all prime ideals of $R$. When $R$ is the ring of polynomials over $\mathbb{C}$ (more generally an algebraically closed field) a remarkable accident happens, and it goes by the name Hilbert Nullstellensatz. This theorem establishes a correspondence between varieties of $\mathbb{C}^n$ and ideals of $\mathbb{C}[z_1,\dotsc,z_n]$. Each variety decomposes into irreducible components and the irreducible components correspond to prime ideals.

As for the localization theorem, I taught this subject a while ago in a graduate course. I only looked at the special case of Hamiltonian $S^1$-actions which already has many nontrivial consequences, and all the ideas needed in the general case are needed in this special as well.

You might want to look over section 3.5 of these course notes where I go through the proof in great detail and at a slower speed than Atiyah-Bott. In the case $n=1$ the question you ask has a more familiar interpretation and the spectrum can be identified with the spectrum of a matrix, whence the the word spectrum. As I point in the notes, the key technical result of the localization theorem was known to A. Borel, almost two decades prior to Atiyah-Bott's work.

Answer by Liviu Nicolaescu for Stratifications and Cohomology Computations

2013-04-24T14:56:05Z

As some guy on the street hinted, any "elementary" inductive approach is a spectral sequence in disguise. You have not indicated what topological invariants you have in mind. If Betti numbers suffices for your needs, then in some instances the spectral sequences become relatively simple.

One such instance comes from the stratification of Grassmannians (or more generally flag manifolds) by Schubert cells. In this case the $k$-th Betti number of the corresponding f;ag space is equal to the number of Schubert cells of dimension $k$.

More generally, suppose that you have a Whitney stratification of a compact space $X$ with the following properties.

All the strata are diffeomorphic to open balls.
There exists no pair of strata $(S,S')$ such that $|\dim S-\dim S'|=1$.

Then the $k$-th Betti number of $X$ is equal to the number of strata of dimension $k$.

Answer by Liviu Nicolaescu for How we do actually compute the topological index in Atiyah-Singer?

2013-04-21T11:44:43Z

As Johannes Ebert said, it's best if at first you stay away from boundary value problems. For some elliptic operators there may not even exist local boundary conditions satisfying the conditions guaranteeing Fredholmness; the Dolbeault operator is such an example. Therefore often one has to deal with pseudo-local boundary value problems such as the Atiyah-Patodi-Singer boundary condition.

A pseudo-diff operator on a closed manifold is Fredholm iff it is elliptic and the index is determined by the principla symbol, which is an element in the $K$-theory of a commutative algebra. For a boundary value problem Fredholmness is a much more subtle issue. It imposes restrictions on the type of boundary value conditions allowed (think Lopatinskii-Schapiro) and as Boutet de Monvel has shown almost four decades ago, the index is determined by the symbol of the problems which is an element in the $K$-theory of a certain non-commutative algebra; see e.g. this paper and the references therein.

The index of an operator on a closed manifold is eminently computable. In most geometric applications it can be reduced to the computation of the indices of a few classical operators: the spin and spin-c Dirac operators, the Hodge-de Rham operator (leading to the Gauss-Bonnet and the Hirzebruch signature operator), Dolbeault operator (leading to the Riemann-Roch-Hirzebruch formula).

The reduction to these cases requires good knowledge of representation theory, differential geometry and extensive familiarity with the theory of characteristic classes.

For manifolds with corners things are even more nebulous; same for most noncompact manifolds. In any case, to paraphrase one of my former professors, if you can describe a PDE problem explicitly, and you can prove its Fredholmness, then the index theorem will give you an answer as explicit as your question.

Answer by Liviu Nicolaescu for How to understand Chern-Simons action

2013-04-19T12:20:55Z

Here is a mid 1970s point of view, courtesy of Atiyah-Patodi-Singer.

Suppose you have a complex vector bundle $E$ of rank $r$ over a smooth manifold $M$. A polynomial function $P$ on the space of $r\times r$ matrices is called invariant if $P(T AT^{-1})=P(A)$ for any $r\times r$ complex matrix $A$ and any invertible $r\times r$ matrix $T$. If you look at

$$ \Delta(x)=\det(1+ xA)=\sum_{k=0}^r c_k(A) x^k, $$

then the coefficient $c_k(A)$ is a homogeneous invariant polynomial function of degree $k$. For example

$$c_1(A)={\rm tr}\; A,\;\;c_r(A)=\det A. $$

To a connection $\nabla$ on $E$ with curvature $F(\nabla)$, we can associate the degree $2k$ form on $E$

$$ c_k(\nabla) = c_k\bigl(\; F(\nabla)\;\bigr), $$

where in the above equality one thinks of $F(\nabla)$ as an $r\times r$-matrix whose entries are $2$-forms. For example

$$ c_1(\nabla)= {\rm tr}\; F(\nabla)= F_{11}(\nabla)+\cdots +F_{rr}(\nabla). $$

Chern-Weil theory proves two things:

The form $c_k(\nabla)$ is closed.
If $\nabla^1,\nabla^0$ are two connections on $E$, then there exists a canonical form of degree (2k-1)$, called the transgression form and denoted by $Tc_k(\nabla^1,\nabla^0)$, which satisfies

$$ d Tc_k(\nabla^1, \nabla^0)= c_k(\nabla^1)-c_k(\nabla^0). $$

In other words, the cohomology class determined by $c_k(\nabla)$ is independent of $\nabla$. This cohomology class is the $k$-th Chern class of $E$.

Suppose now that $\dim M= 2k-1$. Then, on account of dimension, $c_k(\nabla)=0$, yet $Tc_k(\nabla^1,\nabla^0)$ is a top degree form well defined for any choices of $\nabla^0,\nabla^1$.

Suppose additionally that $E$ is trivial and we have fixed a trivialization. Then we can choose $\nabla^0$ to be the trivial connection on $E$ and then we set

$$ CS_k(\nabla):= Tc_k(\nabla,\nabla^0). $$

The usual Chern-Simmons theory is a special case of this construction when $k=2$, i.e., $E$ is a trivial complex vector bundle of rank $r\geq 2$ over a $3$-manifold.

Answer by Liviu Nicolaescu for Characterizing polynomials whose associated semi-algebraic set is bounded?

2013-04-19T14:18:45Z

The question is a bit open-ended because I could give you sufficient and very restrictive conditions for this to happen. Here are a couple of elementary remarks that might help.

Let $[p]$ denote the top degree part of $p$. This is a homogeneous polynomial of degree $k$. If you require that $p>0$ for any $u\in \mathbb{R}^n$ of length $1$ then you get the desired conclusion. Note that this forces the degree $k$ to be even. This reduces the problem to characterizing even degree homogeneous polynomial which are positive definite. There are many of those, but I don't know how to characterize them. Here is a large supply. Suppose that $Q(y_1,\dotsc, y_m)$ is a nontrivial homogeneous polynomial with nonnegative coefficients, and $q_1,\dotsc, q_m$ are positive definite quadratic forms in $n$ variables $x_1,\dotsc, x_n$. Then $Q(q_1,\dotsc, q_m)$ is a positive definite homogeneous polynomial in the variables $x_1,\dotsc, x_n$.

The minimum value of a polynomial $P$ of even degree $k$ on the unit sphere is obtained via Lagrange multipliers. Look at the solutions of the system

$$ \lambda\in\mathbb{R},\;\;u\in\mathbb{R}^n,\;\;|u|=1,\;\; \nabla P(u)=\lambda u. \tag{1}$$

The Euler identity for homogeneous functions implies

$$ k P(u)= u\cdot \nabla P(u) =\lambda |u|^2 =\lambda. $$

The polynomial $P$ will be positive definite if the system (1) has no solutions $(\lambda, u)$ with $\lambda\leq 0$.

Answer by Liviu Nicolaescu for Eigenfunctions of fourth-order differential operator

2013-04-17T15:04:40Z

You need to specify $\lambda_k$ more clearly. More precisely, what is the spectrum of this operator? The equation

$$ \cos x\cosh x =1 $$

seems to have a unique solution $mu_k$ on any interval of the form $(k\pi/2, k\pi/2+\pi)$, $k\in\mathbb{Z}$, so I assume the spectrum might be $\mu_k^4$, $k\in\mathbb{Z}$?!? (Please edit you question to remove this ambiguity.) In any case if the boundary value problem is elliptic (please check that) then the spectrum is discrete. In particular it can be determined by finding the eigenfunctions which means solving some ode's. My guess is that you found all the eigenfunctions, i.e., the system you found is complete.

Update. You need to check two things: 1) the boundary value problem is elliptic 2) it is symmetric. I'll deal with the 2nd issue first because it is faster. Denote by $A$ the operator

$$A=\frac{d^4}{dx^4}. $$

A simple integration by parts shows that for any $u,v\in C^4([0,1])$ we have

$$ \int_0^1 \bigl(\; (Au) -u(Av)\;\bigr) dx=\sum_{j=0}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{j}(0)\;\bigr). $$

If the function $u$ satisfies your boundary conditions $u^{(k)}(x)=0$ for $k=2,3$, $x=0,1$ the above equality simplifies a bit

$$ \int_0^1 \bigl(\; (Au) -u(Av)\;\bigr) dx= \sum_{j=2}^3(-1)^j\bigl( u^{(3-j)}(1)v^{(j)}(1)- u^{(3-j)}(0) v^{j}(0)\;\bigr). $$

If the function $v$ satisfies the same boundary conditions as $u$, then the last equality takes the very simple form

$$ \int_0^1 \bigl(\; (Au) -u(Av)\;\bigr) dx= 0. $$

This says that the boundary value problem is symmetric, or formally selfadjoint.

The ellipticity of this problem is another issue. The most readable account I could find is in Chap. 20 vol.3 of the book The Analysis of Linear Partial Differential Operators by the late great Lars Hormander.

The ellipticity of the boundary value problem requires that the symbol of your operator $A$ be elliptic (which it is) and that the boundary value conditions should satisfy the so called Lopatinskii-Schapiro conditions.

In your case they are trivially satisfied because you work on a one-dimensional space $[0,1]$. The upshot is that in your case the boundary conditions are elliptic. We can form the unbounded operator $\newcommand{\bD}{\boldsymbol{D}}$

$$ \hat{A}: \bD(\hat{A})\subset L^2(0,1)\to L^2(0,1), u\mapsto \frac{d^4 u}{dx^4} $$

Where the domain $\bD(\hat{A})$ of $\hat{A}$ consists of functions in the Sobolev space $L^{4,2}(0,1)$ (four weak derivatives in $L^2$) such that $u^{(j)}(x)=0$ for $x=0,1$, $j=2,3$.

The results in the above monograph show that $\hat{A}$ viewed as an unbounded operator on the Hilbert space $L^2(0,1)$, is closed, densely defined, selfadjoint and has compact resolvent. This is all you need. Arguably, the above argument is a bit heavy, and it feels like hunting a mosquito using a bazooka.

There is a direct, more elementary approach to proving that $\hat{A}$ has compact resolvent. Observe first that the above integration by parts formulae show that the operator $\hat{A}+1$ is positive, i.e.,

$$ (\hat{A}u,u)_2+(u,u)_2>0,\;\;\forall u\in \bD(\hat{A})\setminus 0, $$

where $(-,-)_2$ denotes the $L^2$-inner product. Hence $\hat{A}+1$ is injective. Then follow the strategy in the proof of Theorem 8.22 in Brezis' book Functional Analysis, Sobolev Spaces and Partial Differential Equations to prove that $\hat{A}+1$ is invertible and it's inverse is compact as an operator $L^2(0,1)\to L^2(0,1)$.

Answer by Liviu Nicolaescu for When does one obtain different 3-manifolds by pasting two tori?

2013-04-17T17:51:24Z

Let $M_1, M_2$ be two oriented manifolds with boundary $\newcommand{\pa}{\partial}$ and suppose that $\Phi,\Psi:\pa M_1\to\pa M_2$ are two orientation reversing diffeomorphisms. We get two manifolds

$$ X_\Phi=M_1\cup_\Phi M_2, \;\; X_\Psi= M_1\cup_\psi M_2. $$

If $\Psi^{-1}\circ \Phi: \pa M_1\to \pa M_1$ extends to a diffeomorphism of $M_1$ then $X_\Phi$ and $X_\Psi$ are diffeomorphic.

The answer to your question lies in Kirby calculus. Your manifold seems to have a rather reasonable surgery description, but Kirby calculus is not my cup of tea.

Answer by Liviu Nicolaescu for geometric interpretation of Lie bracket

2013-04-17T10:02:24Z

Arnold liked to call the Lie derivative the "fisherman derivative": you sit on the banks of a river and measure the change in the objects flowing in front of your eyes.

More concretelly, denote by $\Phi^t$ the local flow generated by the vector field $X$. Fix a point $p_0$ in the manifold $M$ and set $p_t:=\Phi^t(p_0)$. Then

$$[X,Y]_{p_0}= \lim_{t\to 0} \frac{1}{t}\Bigl(\;\Phi^{-t}_* Y_{p_t}- Y_{p_0}\;\Bigr)\in T_{p_0}M, $$

where $\Phi^{-t}_*: T_{p_t}M\to T_{p_0} M$ denotes the differential of $\Phi^{-t}$. For a proof I refer to Section 3.1.2 of these lectures.

Answer by Liviu Nicolaescu for Nonharmonic solutions of Laplace's equation

2013-04-16T21:03:28Z

Here is a general result. If $f$ is a distribution on $U$ and $\Delta f\in C^\infty(U)$, then $u\in C^\infty(U)$. More generally, you can replace $\Delta$ with any properly supported elliptic pseudodifferential operator on $U$. In particular, any elliptic operator with smooth coefficients is a properly supported elliptic pseudodifferential operator.

Answer by Liviu Nicolaescu for textbooks on asymptotic expansions

2013-04-16T18:50:18Z

DeBrujin book is an excellent source. Another good source is the litle book Asymptotic Expansions by A. Erdelyi.

Answer by Liviu Nicolaescu for Whitney stratifications

2013-04-11T09:36:16Z

I struggled with the Whitney conditions myself. More precisely, I wanted to understand the geometric significance of these conditions. Apparently Whitney was seeking a simple way to characterize equisingularity: allong a connected component of stratum the stratification looks the same. Technically the Whitney conditions are local conditions that guarantee a desirable global property of the stratification: normal equisingularity.

This turned out to be a difficult problem, first solved by R. Thom, but the first account I could understand would be that by J. Mather.

You can read more about this in Section 4.2 of my book An Invitation to Morse Theory, 2nd edition. There you will find pictures, examples, and the basic geometric consequences of the Whitney conditions, including normal equisingularity. All that in 12 pages and no technical proofs, though rather generous references.

You can find a (less efficient) precursor of Section 4.2 here.

Update In Section 4.3 of the same book I prove that the stratification of a compact manifold by the unstable manifolds of a gradient flow is Whitney iff the flow satisfies the Smale transversality conditions. In particular, the Schubert stratification of Grassmanians is Whitney. For more examples I refer to chapter 7 of this paper.

Answer by Liviu Nicolaescu for Is the space of diffeomorphisms homotopy equivalent to a CW-complex?

2013-04-08T13:40:51Z

Check the paper Antonelli, P. L.; Burghelea, D.; Kahn, P. J. The non-finite homotopy type of some diffeomorphism groups. Topology 11 (1972), 1–49.

There they mention an old result of Palais (Homotopy theory of infinite dimensional manifolds. Topology 5 (1966), 1-16) which states that the identity component of ${\rm Diff}_0(M)$ has the homotopy type of a countable $CW$-complex.

In their paper, Antonelli, Burghelea and Kahn prove that for many smooth manifolds (including spheres of dimension $\geq 7$) the group ${\rm Diff}_0(M)$ does not have the homotopy type of a finite $CW$-complex. (This is highly nontrivial.)

Above, by diffeomorphisms they mean smooth diffeomorphisms and the topology is the $C^\infty$-topology.

Answer by Liviu Nicolaescu for Linear Recurrence Relations in 2 Variables with Variable Coefficients

2013-04-06T17:01:22Z

This recurrence brings to mind the wave equation. Denote by $L$ the lattice $\newcommand{\bZ}{\mathbb{Z}}$ $L=\bZ^2$ and $t:L\to\bZ$ the "time" function

$$ t(x,y)=x+y,\;\; (x,y)\in L. $$

The "initial hypersurface" $S_0$ is defined by the equation $T=0$. A point $p=(x,y)\in L$, $t(p)>0$, is uniquely determined by its time $t(p)=x+y$ and its position $u(p)=x-y$. $\newcommand{\bC}{\mathbb{C}}$

Consider a function $a:L\to\bC$ satisfying your recurrence relation. Its value at the point $(x,y)$ with time $t=(x_0,y_0)$ is only affected by its values in the region $|x-x_0|\leq t$ on initial hypersurface $S_0$.

For any $R>0$ define

$$S_0(R,t)= \lbrace (x,y)\in L;\;\;x+y=t,\;\;|x|\leq R\rbrace, $$

and

$$ m(R, t) :=\max_{(x,y)\in S(R, t)}|a(x,y)|. $$

Your recurrence implies

$$ |a(x,y|\leq \frac{1}{2}\bigl(\; |a(x-1,y)|+|a(x,y)|\;\bigr),\;\;\forall (x,y)\in R. $$

This implies immediately that

$$ m(R,t) \leq m(R+1, t-1). $$

In particular, for $t>0$ we have

$$ m(R,t)\leq m(R+t,0). $$

This controls the size of the "future" values of $a$, i.e., the values of $a$ in the region $t\geq 0$. If we assume that

$$ m(R,0)\leq Ma^R, $$

for some $c\geq 1$, $M>0$ then we deduce that

$$ M(R,t)\leq M a^{R+t},\;\;\forall t\geq 0. $$

In particular, if $a$ is bounded along $S_0$, it will stay bounded in the future. The past values seem a bit more difficult to control. I have to think more about this.

I thought more about this and I reached a conclusion: the past cannot be determined from the initial conditions at $t=0$.

Suppose that we have a function $a: \bZ\to \bC$ satisfying your recurrence conditions and such that, at $t=0$ is zero. What could be the values on the time slice $t=-1$?. Denote by $A$ the restriction of $a$ to the slice $t=-1$. We set $\newcommand{\ii}{\boldsymbol{i}}$

$$ A_n = a(n,-n-1),\;\;n\in\bZ. $$

Set $c:=e^{\ii\theta}$. The recurrence relation implies

$$A_n +c^n A_{n-1}= 0,\forall n\in\bZ $$

so that

$$ A_n = A_0 (-c)^{\ell(n)},\;\;\ell(0)=0,\;\;\ell(n+1)-\ell(n)=n,\;\;\forall n\in\bZ $$

This shows that we can generate solutions of your recurrence that are far from temperate for $t<0$. Here is how you do it.

Fix a function $f_0 :\bZ\to\bC$. This is the initial condition

$$ a(x,-x)=f_0(x),\;\;\forall x\in \bZ. $$

Define $g_{-1}:\bZ\to \bC$ by requiring

$$ g_{-1}(0)=0,\;\;g_{-1}(x)+cg_{-1}(x-1) = 2f_0(x),\;\;\forall x\in \bZ. $$

Pick a constant $M_1>0$ and then set

$$ f_{-1}(x) = M_1 (-c)^{\ell(x)} +g_{-1}(x),\;\;\forall x\in \bZ. $$

Observe that $f_{-1}(0)= M_1+1$ and

$$ f_{-1}(x)+c^xf_{-1}(x-1)=2f_0(x). $$

Proceed inductively. Suppose we have produced functions $f_{-1},\dotsc, f_{-k}:\bZ\to \bC$,

$$ f_{-j}(0)= M_j+1,\;\;j=1,\dotsc, k. $$

We determine $g_{-k-1}:\bZ\to \bC$ by requiring that

$$ g_{-k-1}(0)=0,\;\; g_{-k-1}(x)+c^xg_{-k-1}(x-1)= 2f_{-k}(x),\;\;\forall x\in \bZ. $$

Pick a positive constant $M_{k+1}$ and then set

$$ f_{-k-1}(x) := M_{k+1}(-c)^{\ell(x)}+ g_{-k-1}(x),\;\;\forall x\in\bZ. $$

For $k>0$ we define inductively $f_k:\bZ\to \bC$

$$ f_k(x)=\frac{1}{2}\bigl(\; f_{k-1}(x)+c^xf_{k-1}(x-1)\;\bigr). $$

Finally define $ a:\bZ^2\to\bC $ by setting

$$ a(x,y)= f_{x+y}(x),\;\;\forall (x,y)\in\bZ^2. $$

The function $a$ satisfies the recurrence relation and

$$ a(0, -k) = M_k+1,\;\;\forall k\in \bZ_{>0}. $$

By choosing the sequence $M_k$ suitably, e.g. $M_k = k^{k!}$, your guaranteed a non-temperate behavior for $a$.

Remark 1. Suppose we are given a function $f_0:S_0\to\bC$, a sequence of points $p_n=(x_n,y_n)$, $n>0$, so that $t(p_n)=x_n+y_n)=-n$ and a sequence of complex numbers $C_n$, $n>0$. Then there exists a unique solution $a$ of the recurrence equation satisfying the "initial conditions"

$$ a(p_n)= C_n,\;\;\forall n>0, $$

$$ a(x,y)= f_0(x,y),\;\;\forall (x,y)=S_0. $$

Remark 2. To obtain estimates for the growth of this solutions in the past one needs to understand the growth of the solution of the following initial value problem. Given $f:\bZ\to \bC$ let $u:\bZ\to\bC$ be the solution of the initial value problem

$$ u(x)+c^xu(x-1) = 2f(x),\;\; x\in \bZ, $$

$$u(0) =0. $$

Note that

$$ u(1)= 2f(1),\;\; u(2)= -c^2 u(1) + 2f(2)=-2c^{\ell(2)} f(1)+2f(2), $$

$$ u(3)= -c^3u(2)+2f(3)= 2f(3) -2c^3f(2) +2c^{\ell(3)} f(1) $$

$$ = 2(-c)^{\ell(3)}\bigl(\; (-c)^{-\ell(3)} f(3)+(-c)^{-\ell(2)}f(2)+(-c)^{-\ell(1)} f(1)\;\bigr). \tag{1} $$

The pattern is now clear and one can see that

$$ u(n)\leq 2\bigl( |f(1)|+ \cdots +2|f(n)|\;\bigr). $$

This is an optimal bound which is achieved. Suppose for example that

$$ f(n)= (-c)^{-\ell(n)} r_n,\;\; r_n>0 $$

Then

$$u(n)= 2(-c)^n (r_1+\cdots + r_n), \;\; n>0. $$

The solution with the temperate initial condition

$$ a(n,-n)= (-c)^{\ell(n)} ,\;\;n\in\bZ, $$

$$ a(0,-n)=0,\;\; n>0, $$

is non-temperate because $|a(1,1-t)|=2^t$, $\forall t\geq 0$.

In general is satisfies an estimate of of the type

$$ a(n,n-t)\sim 2^t \frac{n^t}{t!},\;\; t>0,\;\; n>n(t). $$

Remark 3. If the "initial" condition for $a$ is the $\delta$ function concentrated at the orgin, then $a$ has exponential growth in the past.

More precisely, if $a$ satisfies the recurrence and the initial conditions

$$a(x,-x)=0,\;\;\forall x\in\bZ\setminus 0,\;\;a(0,0)=1, $$

$$ a(0,-t)=0,\;\;\forall t<0, $$

then $$|a(1,1-t)|=2^t. $$

Answer by Liviu Nicolaescu for Resolvent of Laplacian

2013-04-04T09:17:20Z

This is mostly enhancing Nik Weaver's comments. Suppose that $M$ is compact of dimension $m$. If $m\geq 2$, then for a generic metric $g$ on $M$ the eigenvalues $\lambda_k$ of the Laplacian $\Delta_g$ are simple. In general, for any $m$, Weyl's spectral estimates imply that

$$\lambda_k \sim C_m \left(\frac{k}{{\rm vol}_g(M)}\right)^{\frac{2}{m}}\;\;\mbox{as $k\to\infty$}, $$

where $C_m$ is an explicit universal constant that depends only on $m$. (Hat-tip to Marc Palm!) In particular this shows that for $m\geq 2$ and a generic metric we have

$$0<\lambda_{k+1}-\lambda_k =O(1). $$

Now Nik Weaver's argument shows that there exists $r_0>0$ such for any $\xi\in [0,\infty)\setminus {\rm spec}\;(\Delta)$ we have $\Vert(R(\xi)\Vert\geq r_0$.

Answer by Liviu Nicolaescu for How many polynomial Morse functions on the sphere?

2013-03-30T12:04:02Z

I've been trying to answer this question for several years and it turned out to be really hard, even for the $2$-sphere. Below I will discuss this case.

First of all one should ask what is the number $m(k)$ of topological types of (stable) Morse functions on $S^2$ with precisely $k$ saddle points. (such a function has $2k+2$ critical points.) I showed that the generating series

$$ x(t) := \sum_{k\geq 0} \frac{m(k)}{(2k+1)!} t^{2k+1}, $$

is the inverse of an elliptic integral; see this paper. More precisely $x(t)$ is the inverse of the function

$$ x\mapsto t(x)=\int_0^x \frac{ds}{\sqrt{s^4/4-s^2-2sx+1}} ds. $$

This fact leads to a positive answer to a question of V.I. Arnold who conjectured that
$$\log m(k)\sim 2k\log k $$

as $k\to \infty $. I refer you to this paper for details. This shows that $m(k)$ grows rather fast as $k\to \infty$.

Any polynomial $P$ of degree $d$ in $\newcommand{\bR}{\mathbb{R}}$ on $\bR^n$ can be uniquely decomposed as a sum

$$ P= \sum_{0\leq j+2k\leq d} r^{2k} H_{j}, \;\; r^2= (x_1^2+\cdots +x_n^2), $$

where $H_{j}$ is a darmonic polynomial of degree $j$. On $\bR^3$ the space of degree $d$ hormonic polynomials has dimension $2d+1$. If we denote by $U_d$ the subspace of $C^\infty(S^2)$ consisting of the restrictions to $S^2$ of the polynomials of degree $\leq d$ we deduce that

$$\dim U_d=\sum_{0\leq k\leq d} (2k+1)=(d+1)^2. $$

Denote by $K_d$ the expected number of critical points of a random function in $U_d$. I showed that

$$ K_d\sim C\dim U_d\sim Cd^2 $$

as $d\to \infty$ where $C$ is a certain explicit constant; see this paper and this paper.

It turns out that the number of critical points of a random function in $U_d$ is highly concentrated around its mean $K_d$, i.e., the probability that the number of critical points of a random function in $U_d$ is far from the mean $K_d$ is extremely small as $d\to\infty$. In more precise technical terms, the variance of the (random) number of critical points of a (random) function in $U_d$ has the same size as $K_d$, which makes the standard deviation of size $\sqrt{K_d}$, much, much smaller than $K_d$ for $d$ large.

I personally believe, based on some empirical evidence, that the mean is close to the maximum number of critical points in the sense that if we denote by $\mu_d$ the maximum number of critical points of a Morse function in $U_d$, then $\mu_d \sim C'' d^2$ as $d\to\infty$.

My guess is that the number of topological types of functions in $U_d$ as $d\to \infty$ is roughly

$$ \sum_{k=1}^{K_d/2} m(k), $$

where I recall that $m(k)$ denotes the number of topological types of Morse functions with $k$ saddle points, i.e., $2k+2$ critical points.

Answer by Liviu Nicolaescu for non-trivial topological line bundles over cartesian product of manifolds not coming from a pullback

2013-03-27T20:25:00Z

Denote by $p_X$ the natural projection $X\times Y\to X$ and define $p_Y$ similarly. Kunneth formula shows that any $\newcommand{\bZ}{\mathbb{Z}}$ $w\in H^1(X\times Y,\bZ/2)$ has the form

$$ w= p_X^* u+p_Y^* v,\;\;u\in H^1(X,\bZ/2),\;\;v\in H^1(Y,\bZ/2). $$

This proves that any real line bundle $L\to X\times Y$ has the form

$$ L= p_X^*L_X\otimes p_Y^*L_Y, $$

with

$$w_1(L)=p_X^*w_1(L_X)+p^*_Y w_1(L_Y). $$

Moreover

$$ L|_{X\times y}= L_X. $$

The line bundles with the properties you want are all pullbacks of nontrivial line bundles on $Y$.

Comment by Liviu Nicolaescu

2013-05-21T11:25:24Z

In the old paper I describe an explicit map $K^{p,q}(X)\to [X,\mathscr{FL}^{p,q}]$. The symplectic reduction is a sort of inverse of this map. Because of space I will give the details as an update to my post.

Comment by Liviu Nicolaescu

2013-05-21T10:06:13Z

I'm with you David.

Comment by Liviu Nicolaescu

2013-05-17T15:38:13Z

Have a look at the book Representations of compact Lie groups by Brocker and tom Dieck, Grad. Texts in Math, vol. 98, Springer.

Comment by Liviu Nicolaescu

2013-05-16T23:59:12Z

Maybe Section 2.1.4 of the notes below might help www3.nd.edu/~lnicolae/ind-thm.pdf

Comment by Liviu Nicolaescu

2013-05-16T15:45:35Z

Good to know. This problem feels so classical that it is not a big surprise.

Comment by Liviu Nicolaescu

2013-05-14T21:15:37Z

Beautiful! Thanks again.

Comment by Liviu Nicolaescu

2013-05-14T17:56:08Z

@George I wish I could click to accept your answer. I like it very much. I was having tunnel vision and your answer broke spell. In my defense, only recently I "discovered" probability, and it seems that I haven't fully warmed up to probabilistic thinking.

Comment by Liviu Nicolaescu

2013-05-14T17:35:31Z

@Duh! Thank you very much. It is embarrassingly obvious now.

Comment by Liviu Nicolaescu

2013-05-14T17:15:32Z

This argument will do the trick for nonsingular $A$'s. In this situation $f$ need not be Lipschitz. For singular $A$'s things are trickier. The measure $\gamma_A$ is the product between the Dirac measure on $\ker A$ and nonsingular Gaussian on $(\ker A)^\perp$.

Comment by Liviu Nicolaescu

2013-05-12T09:28:07Z

Check Klain and Rota's book "Introduction to geometric probability".

Comment by Liviu Nicolaescu

2013-05-09T19:40:54Z

Here is a more recent survey by P. Mironescu of these topics hal.archives-ouvertes.fr/docs/00/74/76/79/PDF/…

Comment by Liviu Nicolaescu

2013-05-09T19:18:40Z

Check section 2.1.4 at this link www3.nd.edu/~lnicolae/ind-thm.pdf These are notes for a graduate course on index theory that I've just taught this semester. Section2.1.4 is about Hodge theory and deals precisely with your question.

Comment by Liviu Nicolaescu

2013-05-09T11:35:43Z

Milnor's book on Morse theory contains applications to a famous second order equations, namely the geodesci equations. Morse's original motivation was investigation of geodescis.

Comment by Liviu Nicolaescu

2013-05-08T20:10:07Z

I found a general formula and it gives me $10$ paths of length four. More details in the update to my post.

Comment by Liviu Nicolaescu

2013-05-07T09:26:54Z

If you add the extra global symmetry $C$, path reversal,then clearly $C^2=1$ and $C$ commutes with the other symmetries $R_x,R_y,J$. The symmetry group has order $16$ and you need to count the number of admissible paths invariant under each of the sixteen symmetries. This looks like an enjoyable and doable problem.

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