User yul otani - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:30:12Z http://mathoverflow.net/feeds/user/21864 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118222/definition-of-operator-valued-integral-with-spectral-measure definition of operator valued integral with spectral measure Yul Otani 2013-01-06T21:06:39Z 2013-01-07T06:01:38Z <p>I am trying to make sense of some operators that come up on Buchholz and Summers' work on warped convolutions (two works on arxiv: <a href="http://arxiv.org/abs/0806.0349" rel="nofollow">2008</a> and <a href="http://arxiv.org/abs/1005.2656" rel="nofollow">2011</a>). </p> <p>There, they work on a Hilbert space $H$ and on the bounded operators algebra $B(H)$ using some operator valued integrals similar to</p> <p><code>$\int_\mathbf{R} A(x)\,dE(x)\;$</code> and <code>$\;\int_\mathbf{R} dE(x)\,A(x)$</code>,</p> <p>where $E$ is the spectral resolution of a self-adjoint operator and $A$ is a $B(H)$ valued (norm-continuous) function. I don't know how one defines that.</p> <p>I don't even know how that one above is defined, since both the measure and the function are operator-valued kinda. What I read about is that you can integrate vector valued functions with respect to a scalar valued measure (Bochner integral or Pettis integral), or scalar valued functions with respect to a spectral resolution (projection valued measure - spectral theorem).</p> <p>If anyone knows how to define it and/or standard references for it, I would be thankful!</p> <hr> <p>Some further remarks. If this helps, the authors also states that if $E$ is the spectral resolution of the self-adjoint operator $P$, then <code>$P\,dE(p) = p\,dE(p)$</code>, and if $B$ is a compact subset of $\mathbf{R}$ and $F$ is a finite-rank projection in $H$, then by spectral calculus <code>$\int_B A(x)\,F\,dE(x)\;$</code> and <code>$\;\int_B dE(x)\,F\,A(x)$</code> are well defined. I can make sense of the second type of integrals, since those will be of finite rank, but not the first type.</p> <p>Thank you.</p> http://mathoverflow.net/questions/114743/is-the-poincare-action-on-the-klein-gordon-quantum-field-strongly-continuous Is the Poincare action on the Klein-Gordon quantum field strongly continuous? Yul Otani 2012-11-28T08:49:09Z 2012-11-28T08:49:09Z <p>I am interested in checking continuity property of the Poincare group action on the Klein-Gordon quantum field theory defined over the Minkowski spacetime. Maybe the simplest example of QFT out there.</p> <p>The setup I am working on is the C*-algebraic one, following Haag's local quantum physics (qft = local net of operator algebras). Let me explain briefly the setup. Take $\mathbf{R} = \mathbf{R}^{1+d}$ the Minkowski spacetime, $P = \partial_t^2 - \Delta + m^2$ the Klein-Gordon differential operator and $G = G^+ - G^-$ the advanced minus retarded propagator. This generates a sympletic vector space $V = C^\infty_c(\mathbf{M}) / P C^\infty_c(\mathbf{M})$ with sympletic form $\sigma([f],[g]) = \int_\mathbf{M} (f Gg) dx$. Now the local net of algebras is the <a href="http://en.wikipedia.org/wiki/CCR_and_CAR_algebras" rel="nofollow">CCR-algebra</a> generated by the Weyl unitaries $W(f)$ for $[f]\in V$, with the rule $W(f)W(g)=e^{-i\sigma(f,g)}W(f+g)$, restricting the support of the test functions to the domain of your local algebra.</p> <p>We have two translation actions on the test functions, which we can call $\alpha$ and $\beta$, such that $\alpha_a f(x) = f(x+a)$ and $\beta_a f(x) = f(x-a)$. Now, we can define the translation action on the C*-algebras by</p> <p>$\alpha_aW(f) = W(\beta_af)$.</p> <p>My question: Is this the right action for the theory? Is this strongly continuous? I think it is not, since for the CCR algebra, it is known that $\|W(f)-W(g)\| = 2$ whenever $[f] \ne [g]$. However, <a href="http://books.google.co.jp/books/about/Local_Quantum_Physics.html?id=a4SIQgAACAAJ&amp;redir_esc=y" rel="nofollow">Haag's book</a> he writes "it is possible and warranted to choose the algebras so that the action of the translation automorphisms on the elements is continuous in the norm topology". What is the problem here?</p> http://mathoverflow.net/questions/101421/distribution-on-lie-groups-and-representations distribution on Lie Groups and representations Yul Otani 2012-07-05T17:51:24Z 2012-07-05T17:51:24Z <p>Let $G$ be a Lie group and $\pi$ a continuous action on $V$ a Fréchet space.</p> <p>This action induces a representation of the compact-supported function $C_c(G)$ (with convolution as product) by</p> <p>$f\in C_c(G) \mapsto \pi(f)\in End(V)$ where <code>$\pi(f)v = \int_G f(g) \pi(g)v \; dg$</code>.</p> <p>On a similar fashion, we can consider the compact-supported bounded Borel measures: $\mu\in M_c(G) \mapsto \pi(\mu)\in End(V)$ where <code>$\pi(\mu)v = \int_G \pi(g)v \; d\mu(g)$</code>.</p> <p>Now my question is: can we define it too to distributions in $G$? If so, is the homomorphism continuous? I ask this because, in the proof of Dixmier-Malliavin theorem, we get some functions $f_n,g \in C^\infty_c(G)$ which $\delta^n * f_n \to \delta + g$ in the sense of distributions ($\delta$ being the Dirac distribution), and then they say that $\pi(\delta^n * f)v \to v + \pi(g)v$ in $V$. How can I prove this?</p> <p>For what I found easily accessible in the literature, $\pi$ on the measures is continuous with regard to the Banach space topology (norm given by the total variation measure), and for $C^\infty_c(G)$, continuity with some $L^1(G)$-norm [That is, <code>$\|{\pi(\mu)v}\|_k \le \| v \|_{n(k)}\int_G |\mu|$</code>, and likewise for $f\in C^\infty_c(G)$.], and that does not seem to help me a lot with distributions... or does?</p> http://mathoverflow.net/questions/100956/some-functional-analysis-questions-laplace-operator-and-fourier-transform/100959#100959 Answer by Yul Otani for Some Functional Analysis Questions (Laplace Operator And Fourier Transform) Yul Otani 2012-06-29T16:58:24Z 2012-06-29T16:58:24Z <p>I cannot comment, so i'll just answer it. I assume you are working on $L^2(R^{2n})$ with Lebesgue measure, right?</p> <p>$\hat{z}_j$ is the operator that multiplies with the $z_j$-coordinate. Any $z \in \mathbf{R}^n$ is written as $(z_1,\ldots,z_j,\ldots,z_n)$, and then we get something like <code>$\big( \, \hat{z}_jf \,\big)(z) = z_j f(z)$</code>. Working with $\Phi$, you can then "put it inside" the integral and derive the equality you mentioned.</p> <p>We didn't get to use anything of the mentioned eigenvalues of this mysterious unnamed operator, though ;)</p> http://mathoverflow.net/questions/98821/how-often-do-people-read-the-work-that-they-cite/100014#100014 Answer by Yul Otani for How often do people read the work that they cite? Yul Otani 2012-06-19T15:21:27Z 2012-06-19T15:21:27Z <p>[offtopic] Since I cannot comment, let me just throw in an old story I heard from my professor. Some time back, a paper by Einstein and Preuss was being cited all around. Now, we all know some names that collaborated with Einsten, but this Preuss is kinda unknown. Turns out that the journal name of the reference</p> <p>Einstein, A. (1931). <em>Sitzungsber. Preuss. Akad. Wiss.</em> ...</p> <p>after some citations, got to be promoted to coauthor. NICE! Here's some (german) reference: <a href="http://de.wikipedia.org/wiki/S._B._Preuss" rel="nofollow">http://de.wikipedia.org/wiki/S._B._Preuss</a></p> <p>Now just to account to the statistics, I know some people that skip the reading of some papers to present seminars and talks, but I think they do check the stuff before writing something up. As to me, I try to read some stuff and then check the references and references of references until I give up. But that doesn't matter, since I'm far from publish anything at all, as it seems. Cheers.</p> http://mathoverflow.net/questions/99860/upper-semicontinuity-in-cx-algebras upper semicontinuity in C(X)-algebras Yul Otani 2012-06-17T20:44:52Z 2012-06-17T22:26:18Z <p>Dear fellows,</p> <p>I've stuck on a step of proposition 1.2 of Rieffel's article (<a href="http://www.springerlink.com/content/t67wk0r6044l3542/" rel="nofollow">continuous field of C*-algebras coming from group cocycles and actions, 1989</a>). I think it basically proves that a C(X)-algebras coming from a locally convex Hausdorff space X is upper semicontinuous, but I can only see that happening for the compact case. The question is simple, but let me elaborate on the background.</p> <p>As definition, of a $C(X)$-algebra, for $X$ compact Hausdorff, it is just a C*-algebra $A$ with a unital <code>$*$</code>-homomorphism $C(X)$ to $ZM(A)$ (center of its multiplier algebra), so it can be though as the Gelfand transform of a C*-subalgebra of $ZM(A)$ containing its identity. Since we can take non-unital C*-subalgebras of $ZM(A)$, it is also interesting to take locally compact but non compact $X$'es. (I think you can check more on that by googling "C(X)-algebras", apparently Kasparov uses this structure in his KK-theory, but I know nothing about that).</p> <p>So let $C_0(X)$ be <code>$*$</code>-isomorphic to a C*-subalgebra of $ZM(A)$. For $x\in X$, take $C_0(X,x)$ as the ideal in $C(X)$ corresponding to functions vanishing at $x$. Then $J_x$ defined as the closure of the linear span generated by $C(X,x)A$ is a closed ideal in $A$. Let $q_x$ be the quotient map from $A \to A_x = A/J_x$. This gives rise to a family of C$^*$-algebras <code>$\{A_x\}_{x\in X}$</code> over $X$.</p> <p>Prop. 1.2 of the reffered paper shows that $A$ is upper semicontinuous. This means that, for all $a\in A$ the map $x \mapsto \| q_x(a) \|_{A_x}$ is upper semicontinuous. This follows from the characterization of the quotient norm.</p> <p>There is a vector of the form $b= \sum_{i=1}^n f_i b_i \in J_x$ (with $f_i(x)=0$ for all $i$) such that $\| a + b\|_A &lt; \| q_x(a) \|_{A(x)}$. Now, since </p> <p>Now, every $f_i$ zeroes at $x$, so we can pick a function $g\in C_0(X)$ such that $g$ is one on a small neighborhood $U$ of $x$ and zero outside a small neighborhood of $U$, and such that $\|g b\|&lt;\epsilon$. Then, for all $y\in U$, we have that $(1-g) \in C(X,y)$ [!] and since $(1-g)f \in J_x$, we get $\| g a \|_A = \| a - (1-g)a \|_A > \|q_y(a)\|_{A_y}$. Thus</p> <p>$\|q_x(a)|_{A_x} > \|g\|\,\|a+b\| - \epsilon > \|ga\| + \|gb\| + \epsilon > \| q_y(a) \|_{A_y} + 2\epsilon.$</p> <p>This proves the upper semicontinuity.</p> <p>NOW HERE IS THE PROBLEM, as we have seen, this is fine for $X$ compact, but if not, I can't see why the constructed $(1-g)a$ is in the ideal $J_y$, since $1$ is an element of the multiplier algebra $M(A)$ and $(1-g)$ would not be an element of $C_0(X)$. Any thoughts on that? Is it true that ANY $C(X)$-algebra is upper semicontinuous then?</p> <p>Also, why should we take, in usual definitions, the $C_0(X)$ is embedded in ZM(A) instead of $C_b(X)$? Otherwise, the constant fields in $C_0(X,A)$ would not be continuous fields... does that any make any sense?</p> http://mathoverflow.net/questions/91104/are-the-smooth-vectors-of-a-frechet-space-dense are the smooth vectors of a Frechet space dense? Yul Otani 2012-03-13T19:25:29Z 2012-03-14T08:10:56Z <p>Given an action $\alpha$ of $V$ a Lie group on $B$ a Fréchet space with seminorms ${ \| \cdot \|_j }$, let $B^\infty$ be the space of smooth vectors. Is this dense in $B$? Can I guarantee it is non-empty? Is there any requirements on $G$ or $\alpha$ or $B$? For the case I am (supposed to be) working on right now, we also have that $\alpha$ is isometric for all the seminorms and strongly continuous. </p> <p>Also, since I am almost illiterate on the subject of Lie groups, I would also be thankful for some easy to read references! Thank you so much.</p> http://mathoverflow.net/questions/125584/measure-space-with-finite-measure Comment by Yul Otani Yul Otani 2013-03-26T03:40:34Z 2013-03-26T03:40:34Z I think Dirac measure would be a counter example. (<a href="http://en.wikipedia.org/wiki/Dirac_measure" rel="nofollow">en.wikipedia.org/wiki/Dirac_measure</a>) http://mathoverflow.net/questions/118222/definition-of-operator-valued-integral-with-spectral-measure/118244#118244 Comment by Yul Otani Yul Otani 2013-01-10T19:14:54Z 2013-01-10T19:14:54Z @Nik, sorry for taking so much time, but thank you for the answer. I'll make remarks here too, as I have been glancing at that paper for some time now. They seem to use the ansatz $\int A(x)dE(x)=\int A(x)dE(y)\delta(x-y)$, and taking $e^{izy}dE(y) = U(z)$ (the generated unitary group), start working the definition for sufficiently regular functions, with some oscillatory integral techniques, mostly with strong operator topology. Rieffel calls those &quot;pseudodifferential operator&quot; techniques. Your definition and example does make sense, of course, but I still have to check how they fit together. http://mathoverflow.net/questions/118222/definition-of-operator-valued-integral-with-spectral-measure Comment by Yul Otani Yul Otani 2013-01-10T19:06:08Z 2013-01-10T19:06:08Z thank you all for your comments! andreas and nik, I think the Riemann sums can be used with the strong operator topology given that the spectral measure $E$ has compact support and the function $A$ has uniform continuity, with maybe some more technical regularity issues. I'm trying to check this, but then again, it seems that it will take quite some time... http://mathoverflow.net/questions/118054/nuclear-space-problem Comment by Yul Otani Yul Otani 2013-01-05T13:24:26Z 2013-01-05T13:24:26Z just to be sure, is $C(X)\otimes A = C(X,A)$ for any C* algebra $A$ (and any C* tensor product)? http://mathoverflow.net/questions/43313/good-references-for-rigged-hilbert-spaces/43506#43506 Comment by Yul Otani Yul Otani 2012-11-14T19:00:03Z 2012-11-14T19:00:03Z Correct me if I'm wrong, but Rieffel's rigged spaces are what we call Hilbert C*-modules, right? Is that related to the rigged spaces everyone here is talking about? I'm asking because I really don't understand... Anyway, if so, a nice book would also be Lance's &quot;Hilbert C*-Modules: A Toolkit for Operator Algebraists&quot;, since its a toolkit and all, very concise and extremely pedagogic. http://mathoverflow.net/questions/103013/probability-of-coin Comment by Yul Otani Yul Otani 2012-07-24T18:56:42Z 2012-07-24T18:56:42Z nice work of google translator or similar, i like the sense of humor: a performer toss the coin in acrobatics practice. nice! http://mathoverflow.net/questions/99860/upper-semicontinuity-in-cx-algebras/99861#99861 Comment by Yul Otani Yul Otani 2012-07-22T20:56:13Z 2012-07-22T20:56:13Z Hello! Thank you again for the comment. I have finally given it some look! I have some considerations. I think that if we evaluate the $x\mapsto \|q_x(a)\|$ defined on $X^*$, the problem is that the ideals and quotients are different to those considering $C_0(X)$ (also, the extension of the injection may not be unique if $C_0(X)$ is degenerate in $ZM(A)$). However, we can simply pick $g$ in $C(X^*)$ instead of in $C_0(X)$, such that all the wanted properties hold, since we only need that $1-g$ to be in $C_0(X)$ (we take the unital extension of the injection $C(X^*)$ in $ZM(A)$). http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/101765#101765 Comment by Yul Otani Yul Otani 2012-07-10T17:38:16Z 2012-07-10T17:38:16Z Borges is too good, a must read! &quot;the library of Babel&quot; often gets me to think about infinity, since the problem posed there was in fact finite, but &quot;sensorially infinite&quot;. http://mathoverflow.net/questions/101420/music-mathematical-point-of-view-revised Comment by Yul Otani Yul Otani 2012-07-05T18:01:59Z 2012-07-05T18:01:59Z I have no idea of what it says, but there is a paper called &quot;Towards A Categorical Approach of Transformational Music Theory&quot; by Alexandre Popoff [<a href="http://arxiv.org/abs/1204.3216]" rel="nofollow">arxiv.org/abs/1204.3216]</a>. I saw this at John Baez' post <a href="https://plus.google.com/u/0/117663015413546257905/posts/fLByuSqNew9" rel="nofollow">plus.google.com/u/0/117663015413546257905/posts/&hellip;</a>. http://mathoverflow.net/questions/100956/some-functional-analysis-questions-laplace-operator-and-fourier-transform/100959#100959 Comment by Yul Otani Yul Otani 2012-06-29T17:36:40Z 2012-06-29T17:36:40Z I'm not sure of what you mean... &quot;derive&quot; in the sense of derivative, in the sense motivating the appearance of the formula, or in the sense of getting from the RHS to the LHS? If so, on which step? Can you understand it by going backwards? For instance, $\frac{\partial}{\partial x_j} e^{i x z} = \frac{\partial}{\partial x_j} e^{i x_1 z_1}e^{i x_2 z_2}\ldots e^{i x_n z_n} = iz_j e^{i x z}$. Just a matter of understanding that $x z$ means the usual inner product in $\mathbf{R}^n$. http://mathoverflow.net/questions/99860/upper-semicontinuity-in-cx-algebras Comment by Yul Otani Yul Otani 2012-06-18T07:17:21Z 2012-06-18T07:17:21Z Hello Mr. Majer, thank you for your comment. Seems like I wasn't clear enough in the definitions. For any point $x$ of $X$, I call $C_0(X,x)$ the (closed ideal) set of (complex valued) function $f\in C_0(X)$ such that $f(x)=0$ (at the particularly defined $x$). This is, of course, not equal to $C_0(X)$. If we take $C(X,x)$ in the same fashion, as the set of functions $f\in C(X)$ such that $f(x)=0$, then $C_0(X,x)$ and $C(X,x)$ are equal iff $X$ is compact. Thank you. http://mathoverflow.net/questions/99860/upper-semicontinuity-in-cx-algebras/99861#99861 Comment by Yul Otani Yul Otani 2012-06-18T07:12:25Z 2012-06-18T07:12:25Z Dear Mr. Weaver, thank you for your help! I'll dig in the proof of what you said and look for basic references, since I am illiterate on the subject :( Anyways, if you have other comments on to why take C_0 and not C_b, I should pay attention! Cheers and thanks again!