User pradip mishra - MathOverflow

Answer by Pradip Mishra for Loop space: De Rham cohomology

2012-08-23T19:53:33Z

This is comment rather than answer: Please check it, whether it makes sense...

Corollary 2.6 page 11 of Free Loop space and homology by J.L Loday says that For any simply connected space, there is a functorial isomorphism: $$HH_1 (\Omega^1(M)) \cong H^1(LM)$$ And Hochschild-Kostant-Rosenberg theorem says that: For a k-algebra $R$, its module of Kähler differentials coincides with its first Hochschild homology $$\Omega_1(R/k)\cong HH_1(R)$$

Now we have by this MO post, a surjective map $\Omega_1(C^\infty(M))\to \Omega^1(M)$.

So can we say that $H^1(LM)= \Omega_1(C^\infty(M))$ and if $\Omega^1(M)\neq {0}$, we have $H^1(LM)\neq {0}$ for simply connected finite dimension manifold $M$.

Analytic extension across the boundary.

2012-05-22T17:31:02Z

Let $Q=[0,\infty)\times [0,\infty)\subset \mathbb C$ and $f: Q\times Q\to Q\times Q$ be a diffeomorphism. such that $f$ is holomorphic in the interior of $Q\times Q$. Can we extend this map analytically across the boundary.

Motivation: We have following proposition:

Let $U$ and $V$ are open subsets of $\mathbb R^n_k=[0,\infty)^k\times \mathbb R^{n-k}$ and $f:U\to V$ be diffeomorphism, then

(a). $x\notin \partial U \Leftrightarrow f(x)\notin \partial V$

(b). $f|Int(U)$, and $f|\partial U$ are diffeomorphism.

This proposition gives: If $f:Q\to Q$ is diffeomorphism and holomorphic in the interior. Then either

1- $f$ maps Y-axis to Y-axis and X-axis to X-axis origin goes to origin. OR

2-$f$ maps X-axis to Y-axis and Y-axis to X-axis origin goes to origin.

And using Schwarz reflection principle we have extension in both case. So for $f: Q\to Q$ we have extension across the boundary. I have doubt for $f:Q\times Q\to Q\times Q$.

hodographic transformation

2012-04-04T08:56:01Z

Let $\phi(x,t)$ be smooth function.

Let $\zeta= x-t$ and $\eta= x+t$. $u=\phi_\zeta$, $v= \phi_\eta$.

Let $u$, $v$ satisfies following equations:

1- $$u_\eta- v_\zeta= 0$$ $$v^2u_\zeta-(1+2uv)u_\eta+ u^2v_\eta=0$$

The roles of dependent and independent variables are then interchanged to give

2-$$\zeta_v-\eta_u=0$$ $$v^2\eta_v+(1+2uv)\zeta_v+u^2\zeta_u=0$$

Can someone please help me to find the required transformation for getting set 2 from set-1. Is this an example of hodographic transformation. Can someone be precise in explaining this transformation.

A corollary to Stone-Weierstrass theorem

2012-02-08T06:01:41Z

Can i get the answer to the following problem. I am having a proof, i feel there is something wrong here..Can you please point out!

Let $D\subset \mathbb C$ be a simply connected domain, and $\gamma: [0,1]\to D$, be a smooth embedding. Given a continuous one form $\phi$ along $\gamma$ and $\epsilon >0$, Does there exists a holomorphic function $h$ on some open neighborhood $U$ of $\gamma$, $U\subset D$ such that $|dh-\phi|<\epsilon$.

Suggested Proof:

Without loss of generality we can assume that $0\notin D$. We can write $\phi= \phi_1 d\zeta$, with $\phi_1$ a continuous function on $\gamma$. We can uniformly approximate $\phi_1$ by Laurent polynomials of the form $\psi_1(\zeta)= \sum_{-k}^k a_j\zeta^k$. As $0\notin D$, we have $\psi_1(\zeta)$ analytic on some possibly small simply connected subdomain of $D$ which we will denote by $D$ itself.

We know that if D is a simply connected domain and $\psi_1$ is analytic in D, then $\psi_1$ has an antiderivative at all points of D. Hence take $h(z)= \int \psi_1(\zeta)$ which will be our required holomorphic function.

Isometric Immersion of $S^1\to M$

2012-03-15T09:39:45Z

$M$ be any Riemannian manifold, and $S^1$ is a circle. We can give Manifold structure to $C^\infty(S^1, M)$ modeled on nuclear frechet space.

Take $Imm(S^1, M):\{f\in C^\infty(S^1,M): f \text{ is an immersion} \}$ . This is open subset of $C^\infty(S^1,M)$ hence it is also a nuclear frechet manifold.

Consider set of all isomoetric immersion of $S^1\to M$, can we give some differentiable structure here. Please provide the reference where people have already studied the isometric immersed loops over a manifold.

Edit: Can we see Set of isometric immersion as a manifold modeled over some Locally convex space.

What is soliton

2012-03-05T18:08:15Z

I am new to this word.. This is not research level problem and it is soft question in nature. Just for curiosity, i am asking..

In literature, i am finding following words:(Wikipedia+ others).

Soliton is a self-reinforcing solitary wave

Solition is a phenomenon.

Solition is a property

Solitonic solution

As wikipedia also says, single definition is difficult to find. Can somebody explain this term according to you... It will be better if you give the idea of Soliton more mathematically rather only intuitively.

More precisely my question is WHAT IS SOLITON.

Answer by Pradip Mishra for What out-of-print books would you like to see re-printed?

2012-02-19T09:38:55Z

Manifolds of differential mapping: P.W. Michor. should be printed with latex and graphics...

What does non-levi flat point mean geometrically

2012-02-17T22:21:51Z

Hello,

$CR$ manifold for example $S^1\times C^{n-1}$ is every where levi flat. Can I have example of $CR$ manifold which has at least one non levi flat point.

I can't see what the happening in Non-Levi flat points. Sorry for vague question and if it is trivial.. .... but basically i want to understand the non levi flat point such that i can easily determine all non levi flat point of given CR manifold...

Any comment and suggestion are welcome.. Thanks in advance.

Answer by Pradip Mishra for On the smooth structure of the spaces of $k$-jets

2012-02-09T06:35:21Z

Let $(U,u)$ is a chart for $M$, and $(V,v)$ be a chart for $N$. $u: U\to u(U)\mathbb R^n$ is diffeomorphism. $u(U)$ and $v(V)$ are open subset of $\mathbb R^n$ and $\mathbb R^m$. Then we can identify $$J^k(u(U),v(V))= u(U)\times v(V)\times \Pi_{j=1}^k L^j_{sym}(\mathbb R^n, \mathbb R^m)$$ People give manifold structure on $J^K(M,N)$ by chart $(J^k(U,V), J^k(u^{-1}, v))$. Main aim is to define map $J^k(u^{-1}, v)$. $$J^k(u^{-1}, v): J^k(U,V)\to J^k(u(U), v(V))\text{ is defined as following:}$$

Firstly for $u:U\to u(U)$ define $J^k(u,V):J^k(U,V)\to J^k(u(U),V)$ by $ J^k(u,V)j^kf_x= j^k(fog)_{g^{-1}(x)}$. This is a well defined map and $J^k(u,V)^{-1}= J^k(u^{-1},V)$

Now same way for $v$ define map $J^k(U,v): J^k(U,V)\to J^k(U, v(V)$. Take $$J^k(u^{-1}, v):= J^k(u^{-1},v)oJ^k(u(U),v)$$ This will be bijective and satisfy coordinate transformation condition:

For details please see: First Chapter 1.1 to 1.8 of Manifolds of differential mapping: P.W. Michor.

Answer by Pradip Mishra for On the determination of a quadratic form from its isotropy group

2012-02-07T04:20:18Z

If $v=0$, then we have $F(v)= G(v)=0$ Hence obviously we have $F(v)= \lambda. G(v)$ for some non zero real $\lambda$.

Now if $v\neq 0$, then for any $g\in GL_n(\mathbb R)$, we have $g(v)\neq 0$ and $g^{-1}(v)\neq 0$ and hence $F(g^{-1}(v))\neq 0$ and $G(g^{-1}(v))\neq 0$, as $F$ and $G$ are non-degenerate.

If $g\in O(F)$ we have $g\in O(G)$. We have $\forall v\in \mathbb R^n, v\neq 0$,

$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$ $$F(g(v)). G(v)= F(v). G(g(v))$$ $$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$ $$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$

Edited: But it doesn't says that $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ is independent of $v$. So proof is incomplete.

every where levi flat

2012-01-08T10:49:43Z

"Suppose $N$ is $2n-1; n\geq 2$ dimensional $CR$ manifold and everywhere Levi flat, then it will be locally $CR$ equivalent to $S^1\times \mathbb C^{n-1}.$"

Above statement can be found in Loop space as complex manifolds paper by Laszlo Lempert, J. Differential geometry 38(93),519-543.http://intlpress.com/JDG/archive/1993/38-3-519.pdf. But I am not able to see it. May be i didn't get the meaning of Levi flat properly..

Can someone please provide the reference material to understand much about Levi flat and Non levi flat point manifolds.

Where the things are wrong in the following example:

Take $N= (0,1)\times \mathbb C\equiv (0,1)\times \mathbb R^2.$ Let $t, x, y$ be coordinate. For $p\in N$, $T_pN= \mathbb R \{\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}$ . Define $H_p^{1,0}N= \{a.\frac{\partial}{\partial z}: a\in \mathbb C\}$ . Then $H_p^{1,0}N$ defines $CR$ structure on $N$. And with this structure, $N$ is every where Levi flat.

Answer by Pradip Mishra for Covariant derivative

2012-01-05T05:50:36Z

I don't know much about infinite dimensional things. I am not sure about the right answer but may be following may be useful... Once i saw the following book and statement:

See Page 4 of the book "Lectures on closed geodesics"- W. Klingenberg. Where he says:

"Whereas for Euclidean vector bundles over Euclidean manifolds such a map (Covariant derivative) $\nabla $ always defines a connection $K$, in our more general situation (That is Loop space: Hilbert Manifold) this need not always be true; see [FK] for further details. See also [El 3] for a more general setting."\

EL3: Eliasson, H.: On the geometry of manifolds of maps. J. Diff. Geom. 1, 165 -194 (1967).

So as far as I know in infinite dimension we can define co variant derivative which doesn't come from any so called connection.

Conformal Extension from a closed set to open

2011-09-17T21:38:20Z

Let $Q = \{(x,y): x,y\geq 0\} $ be the 1st quadrant of $\mathbb R^2$, and $f$ is a function defined on it such that all the partial derivative(any order) of $f$ exists and continuous. By Whitney extension theorem (1934 Proceedings of AMS) we know there exists a functions $\tilde{f}$ and open set $\tilde{U}$, such that $\tilde{f}$ is defined on $\tilde{U}$ and $\tilde{f}$ is smooth, $Q\subset \tilde{U}$ and $\tilde{f}$ restricted to $Q$ is $f$. Also $\tilde{f}$ is not unique but for each $p\in Q$, $d\tilde{f}_p$ is same. Now define $df_p:= d\tilde{f}_p$

Now for each $p\in Q$ , let $R(p)$ be a constant time rotation matrix at $p$, This mean $R(p)= c(p) \left[ {\begin{array}{cc} \cos \theta(p) & \sin \theta(p),\ -\sin \theta(p) & \cos \theta(p)
\end{array} } \right]$

$R(p)$ is $2\times 2$ matrix (pls someone fix the tex), and $c(p)$ is differentiable function on $Q$ to $\mathbb R$. and we have for each $p\in Q$, $df_p= R(p)$, does there exits any $\tilde{f}$ and $\tilde{U}$ as above such that for each $q\in \tilde{U}$ we have $d\tilde{f}_q= S(q)$. Where $S(q)$ is constant time rotational matrix and $S(p)= R(p)$ for each $p\in Q$.

Conformal extension

2011-08-30T13:28:00Z

Does there exist a conformal smooth extension of a smooth function? Smooth extension is guaranteed by Whitney extension theorem. does that theorem also says for conformality. Precisely the question is the following:

Let $ D= \{z: |z|\leq 1,I[z]\geq 0 \}$ . Define $f\colon D \to\mathbb R^2$ such that all the derivative of $f$ exists and continuous and for all $x\in D$ and $u,v\in\mathbb R^2$, we have $\langle f'(x)(u) ,f'(x)(v)\rangle=\alpha (x)\langle u,v\rangle$, where $\alpha $ is some positive function on D.

Now does there exists an open set $U$ in $\mathbb R^2$ such that $D\subset U$, and a smooth function $g$ on $U$ which restricted to $D$ is $f$ and $\langle g'(x)(u),g'(x)(v)\rangle=\beta(x)\langle u,v\rangle$ for all $x \in U$ for $\beta $ some positive function on $U$.

Is it possible to see Path Spaces as manifold

2011-06-27T06:46:09Z

Dear Sir/friends,

How to give manifold structure to set of all $C^2$ path over any manifold.

Comment by Pradip Mishra

2013-01-18T15:00:58Z

I mean for each $t$, not as function... but it is true that if $\gamma'(t)=0$ then it will be zero... but then for my application in mind, things wouldn't work.. I think, i should give some more thought, what i need. Thanks a lot for the comment.

Comment by Pradip Mishra

2013-01-02T11:53:19Z

Can someone please help me.. please make some comment/or give answer....Feel free to retag if needed.. Is this question too trivial or not well written. Thanks

Comment by Pradip Mishra

2012-05-30T08:16:21Z

I started learning linear algebra in my Undergraduate days from Linear Algebra (2nd Edition)" by Kenneth M Hoffman and Ray Kunze. I recommend this.

Comment by Pradip Mishra

2012-05-23T16:12:02Z

This argument is valid for map(diffeomorphism and in interior holomorphic) $f: Q\times H\to Q\times H$ where $H$ is upper half plane... Nothing special about $Q$.. am i right?

Comment by Pradip Mishra

2012-05-23T16:03:25Z

thanks for the answer.

Comment by Pradip Mishra

2012-04-05T02:14:03Z

@Robert Bryant sir, Thanks for the giving a nice interpretation.

Comment by Pradip Mishra

2012-04-05T02:12:15Z

@Wilie wong, Thanks a lot for explaining.

Comment by Pradip Mishra

2012-03-20T06:21:49Z

If $\eta$ is a section of $T^{0,1}M$ then statement is trivial...

Comment by Pradip Mishra

2012-03-20T06:20:35Z

Yes reference is: Lempert: Loop space as complex manifolds, Journal of differential geometry,38, 1993

Comment by Pradip Mishra

2012-03-19T15:02:08Z

ohh sorry.. i got.. and thanks for the time...

Comment by Pradip Mishra

2012-03-19T11:52:30Z

May be i am making mistake.. Please have the following example: Take K any smooth curve which doesn't passes through $(0,0)∈\mathbb C$. Take $f(z)=\frac{1}{z}$, on K, f is continuous, and $\mathbb C−K$ is connected. But f can't be extended to a entire function.... So the theorem you mention seems to have some problem.

Comment by Pradip Mishra

2012-03-16T05:22:12Z

@Arch Stanton, and @Deane Yang.... Where the proof is wrong... I can't see... If we fix a metric of $h$ on $S^1$... then there is a diffeomorphism $\phi: S^1\to S^1$ such that induced metric from some immersion $f^*g$ is same as $h$ via $\phi$. Notation is same as in answer.

Comment by Pradip Mishra

2012-03-16T04:34:58Z

thanks for the references.

Comment by Pradip Mishra

2012-03-15T12:03:11Z

Edited.. Differentiable structure: I mean manifold structure.

Comment by Pradip Mishra

2012-03-15T12:02:37Z

@Andew Stacey... So if i understand correctly, You are trying to give manifold structure to set of all immersion....