User branimir ćaćić - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:49:53Z http://mathoverflow.net/feeds/user/6999 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122748/what-is-a-spinor-structure/122768#122768 Answer by Branimir Ćaćić for what is a spinor structure? Branimir Ćaćić 2013-02-24T01:14:21Z 2013-02-25T22:01:48Z <p>Just to elaborate a bit in explicitly differential-geometric terms on MTS's answer, which refers to <a href="http://www.theta.ro/jot/archive/1986-016-002/1986-016-002-008.pdf" rel="nofollow">certain results of Plymen's</a> originally restated in terms of Morita equivalence (via the dictionary given by the Serre–Swan theorem), let $M$ be a compact orientable Riemannian manifold, and let $\operatorname{\mathbb{C}l}^{(+)}(M)$ be the finite rank Azumaya bundle given by the complexification of the Clifford bundle $\operatorname{Cl}(M)$ if $\dim M$ is even, and by the complexification of the even subbundle of the Clifford bundle if $\dim M$ is odd. Then $M$ is <em>spin</em>$^\mathbb{C}$ if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module, that is, a Hermitian vector bundle $\mathcal{S} \to M$ (i.e., <em>a</em> spinor bundle) such that $\operatorname{\mathbb{C}l}^{(+)}(M) \cong \operatorname{End}(\mathcal{S})$.</p> <p>Now, if what you care about are specifically <em>spin</em> manifolds, one can endow $\operatorname{\mathbb{C}l}^{(+)}(M)$ with a canonical $\mathbb{C}$-linear anti-involution, and hence equip the dual bundle $\mathcal{E}^*$ of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module with the structure of a $\operatorname{\mathbb{C}l}^{(+)}(M)$-module. It is then , that $M$ is actually <em>spin</em> if and only if there exists an irreducible $\operatorname{\mathbb{C}l}^{(+)}(M)$-module $\mathcal{S} \to M$ such that $\mathcal{S} \cong \mathcal{S}^\ast$ not only as Hermitian vector bundles, but also as $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules—this is, then, <em>the</em> spinor bundle for its corresponding spin structure. Indeed, by the anti-unitary isomorphism $\mathcal{S} \cong \mathcal{S}^\ast$ of Hermitian vector bundles defined by the Hermitian metric, together with a little bit of care, one can recognise such a unitary isomorphism of $\operatorname{\mathbb{C}l}^{(+)}(M)$-modules as nothing else than the charge conjugation operator on spinors in mild disguise.</p> http://mathoverflow.net/questions/118469/hopf-algebra-for-a-physicist/118470#118470 Answer by Branimir Ćaćić for Hopf Algebra for a physicist Branimir Ćaćić 2013-01-09T17:55:02Z 2013-01-09T17:55:02Z <p>Dominique Manchon's <a href="http://arxiv.org/abs/math/0408405" rel="nofollow">lecture notes</a>, which are very well-known amongst people working on Connes--Kreimer renormalisation, offer exactly the sort of detailed, accessible introduction to Hopf algebras and Connes--Kreimer renormalisation that you're looking for. However, you should first be thoroughly comfortable with abstract linear algebra and with the basics of ring and module theory, and you should be familiar with the basic language of category theory and of representation theory. Roughly speaking, if you can follow along with Chapter 1 of the notes, you should have the bare minimum needed.</p> http://mathoverflow.net/questions/118396/elaborating-mercers-theorem-rkhs-on-cameron-martin-space-kx-y-minx-y/118414#118414 Answer by Branimir Ćaćić for Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$ Branimir Ćaćić 2013-01-09T01:02:09Z 2013-01-09T01:46:33Z <p>If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then $$\mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \alpha x\sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \alpha x \cos(\alpha T))$$ (up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is $$\psi_k(x) = \sqrt{\frac{2}{T}} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right)$$ with corresponding eigenvalues $$\lambda_k = \frac{T^2}{\pi^2(k+\tfrac{1}{2})^2}.$$ Hence, $$k(x,y) = \sum_{k=0}^\infty \frac{2T}{\pi^2(k+\tfrac{1}{2})^2} \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}x\right) \sin\left(\frac{\pi(k+\tfrac{1}{2})}{T}y\right).$$ I hope this works!</p> http://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds/117813#117813 Answer by Branimir Ćaćić for Noncommutative smooth manifolds Branimir Ćaćić 2013-01-01T21:59:16Z 2013-01-01T21:59:16Z <p>I'm a bit wary of resurrecting such an old question, but given that the precise content of the reconstruction theorem doesn't seem to be terribly well disseminated, please permit me to <a href="http://math.stackexchange.com/a/268334/49610" rel="nofollow">cross-post from math.SE</a> and then make some extra comments:</p> <p>"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:</p> <ul> <li>A unital Frechet pre-$C^\ast$-algebra $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact orientable $p$-manifold if and only if there exists a $\ast$-representation of $A$ on a Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$.</li> <li>In particular, $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact <em>spin$^{\mathbb{C}}$</em> $p$-manifold if and only if there exist $H$ and $D$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$ <em>and</em> $A^{\prime\prime}$ acts on $H$ with multiplicity $2^{\lfloor p/2\rfloor}$.</li> </ul> <p>"Once you know that $A \cong C^\infty(X)$, you can then apply the much earlier "baby reconstruction theorem" (for lack of a better phrase) announced by Connes and proved in detail by Gracia-Bondia--Varilly--Figueroa to conclude that:</p> <ul> <li>In the general case, $(A,H,D) \cong (C^\infty(X),L^2(X,E),D)$ where $E \to X$ is a Hermitian vector bundle and $D$ can be interpreted as an essentially self-adjoint elliptic first-order differential operator on $E$.</li> <li>In the case where $A^{\prime\prime}$ acts with multiplicity $2^{\lfloor p/2 \rfloor}$, $E \to X$ is in fact a spinor bundle (i.e., irreducible Clifford module bundle) and $D$ is Dirac-type (viz, a perturbation of a spin$^{\mathbb{C}}$ Dirac operator by a symmetric bundle endomorphism of $E$).</li> </ul> <p>"So, whilst you can refine the reconstruction theorem to a characterisation of compact spin$^{\mathbb{C}}$ manifolds with spinor bundle and essentially self-adjoint Dirac-type operator, the general result is really just a statement about compact orientable manifolds. Indeed, one can even refine the reconstruction theorem to a characterisation of compact oriented Riemannian manifolds with self-adjoint Clifford module and essentially self-adjoint Dirac-type operator."</p> <p>As for why you need more than just an algebra $A$, here's what I understand of the situation:</p> <ol> <li>Gel'fand--Naimark says that a commutative unital $C^\ast$-algebra gives a compact Hausdorff space, no more and no less.</li> <li>Going by the example of $C^\infty(X) \subset C(X)$, one might try considering commutative unital Frechet pre-$C^\ast$-algebras, but one can readily cook up examples of such algebras that <em>aren't</em> isomorphic to $C^\infty(X)$ for some $X$. So, if one still wishes to follow this line of inquiry, one would need still more structure.</li> <li><p>In terms of the reconstruction theorem itself (which does tend to be treated as a black box), Connes's proof (insofar as I can understand) really makes absolutely essential use of all three parts of the spectral triple $(A,H,D)$:</p> <ul> <li>the norm closure of $A$ in $B(H)$, by Gel'fand--Naimark, gives you a (canonical) compact Hausdorff space $X$;</li> <li>the noncommutative integral defined by $D$ gives a Radon measure on $X$;</li> <li>the Hochschild cycle of the orientability condition gives you candidates for charts;</li> <li>the operator $D$ then gets used to show that you actually have a smooth atlas. </li> </ul> <p>This doesn't suggest, of course, that a spectral triple is the optimal notion of algebra + extra data qua noncommutative manifold, but it does suggest that it might well be a reasonably economical definition. In particular, it suggests that the real puzzle with the spectral triple formalism isn't the extra Riemannian data, but rather the seemingly absolute necessity of orientability.</p></li> </ol> <p>My apologies for the long-windedness!</p> http://mathoverflow.net/questions/117156/morita-equivalence-for-algebras/117159#117159 Answer by Branimir Ćaćić for Morita equivalence for *-algebras Branimir Ćaćić 2012-12-24T17:58:26Z 2012-12-24T18:11:08Z <p>Perhaps what you want is something along the lines of Rieffel's strong Morita equivalence of $C^\ast$-algebras (see, for instance, <a href="http://math.berkeley.edu/~alanw/242papers99/bursztyn" rel="nofollow">http://math.berkeley.edu/~alanw/242papers99/bursztyn</a>, and pretty much any introductory account of operator-algebraic noncommutative geometry), except forgetting all the functional-analytic nuances?</p> http://mathoverflow.net/questions/115553/index-theorems-and-orientability/115571#115571 Answer by Branimir Ćaćić for Index theorems and orientability Branimir Ćaćić 2012-12-06T03:45:01Z 2012-12-06T03:45:01Z <p>The trick, rather, is to write $\text{Ind}(D)$ in terms of the (co-)homology of (the total space of) $T^\ast M$, which <em>is</em> always orientable, viz, $$\text{Ind}(D) = \int_{T^\ast M} \text{ch}[\sigma_m(D)] \smile \text{Td}(T^\ast M \otimes \mathbb{C})$$ where $\text{ch}[\sigma_m(D)] \in H^{\text{even}}(T^\ast M)$ is the Chern character of the symbol class $\sigma_M(D) \in K(T^\ast M)$ of $D$ and $\text{Td}(T^\ast M \otimes \mathbb{C})$ is the Todd genus of $T^\ast M \otimes \mathbb{C}$. For an explanation, see </p> <p><a href="http://mathoverflow.net/questions/23409/intuitive-explanation-for-the-atiyah-singer-index-theorem" rel="nofollow">http://mathoverflow.net/questions/23409/intuitive-explanation-for-the-atiyah-singer-index-theorem</a> </p> <p>and especially Paul Siegel's excellent answer. In particular, written this way, the Atiyah--Singer index theorem can be viewed as the translation into (co-)homological terms of a purely $K$-theoretic statement.</p> http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator/114202#114202 Answer by Branimir Ćaćić for Integral kernel for the resolvent of the laplace operator Branimir Ćaćić 2012-11-23T01:31:59Z 2012-11-23T08:05:11Z <p>At least when $\Re z &lt; 0$, and assuming that you're taking $\Delta$ to be a negative operator (i.e., $-\Delta \geq 0$), you can write $(s-z)^{-1} = \int_0^\infty e^{zt} e^{-st} dt$, so that by the functional calculus, you should be able to write your resolvent as $Ru = \int_0^\infty e^{zt} e^{\Delta t}u dt$, and hence your integral kernel as $$K(x,y) = \int_0^\infty e^{zt} K_H(x,y;t) dt,$$ where $$K_H(x,y;t) = \frac{1}{4\pi t}e^{-|x-y|^2/4t}$$ is the heat kernel for $\mathbb{R}^2$. Even if this sketch is complete nonsense, I still suspect that heat kernel methods might nonetheless be helpful, at least in the regime $\Re z &lt; 0$.</p> <p>EDIT: Helpful perhaps with regards to any estimates you might want--not so helpful with regards to evaluating your perfectly concrete integral...</p> http://mathoverflow.net/questions/110113/well-defined-tensoring-of-spectral-triples/112544#112544 Answer by Branimir Ćaćić for Well defined Tensoring of spectral triples Branimir Ćaćić 2012-11-16T02:26:44Z 2012-11-16T02:33:25Z <p>There's a reasonably convenient abuse of notation in play. When dealing with almost-commutative spectral triples, $C^\infty(M)$ means $C^\infty(M,\mathbb{C})$ except when forming $C^\infty(M) \otimes A_F$ for $A_F$ real, in which case $C^\infty(M) \otimes A_F$ really means $C^\infty(M,\mathbb{R}) \otimes_{\mathbb{R}} A_F$.</p> http://mathoverflow.net/questions/119898/when-is-a-pseudo-differential-operator-trace-class-or-in-dixmier-ideal Comment by Branimir Ćaćić Branimir Ćaćić 2013-01-28T10:30:25Z 2013-01-28T10:30:25Z Moreover, by the Connes trace formula (<a href="http://www.alainconnes.org/docs/action88.pdf" rel="nofollow">alainconnes.org/docs/action88.pdf</a>), if your operator is of order $-k$ for $k = \dim M$, then it is in the Dixmier ideal; indeed, it is measurable (in the sense of the theory of Dixmier traces), and the (unique value of the) Dixmier trace is given by the Wodzicki residue of your operator. http://mathoverflow.net/questions/118206/eigenvalues-of-the-free-sphere Comment by Branimir Ćaćić Branimir Ćaćić 2013-01-08T01:59:09Z 2013-01-08T01:59:09Z @Z254R Is $\sqrt{d^\ast d}$ actually going to give you the &quot;Dirac operator&quot; of a spectral triple? From the look of it, I'd sooner expect $\sqrt{d^\ast d}$ to only be the absolute value of such an operator, and finding the correct &quot;sign&quot; is often the tricky part with constructing spectral triples from the ground up. http://mathoverflow.net/questions/118226/seeing-topological-geom-properties-of-the-space-via-corresponding-c-algebra/118252#118252 Comment by Branimir Ćaćić Branimir Ćaćić 2013-01-07T19:05:43Z 2013-01-07T19:05:43Z @IgorKhavkine: Would you happen to know how the approach of Michor and Vanžura compares to that outlined in &quot;Smooth Manifolds and Observables&quot; by &quot;Jet Nestruev&quot;? They seem to be at least in the same spirit, though I don't know how the details line up. http://mathoverflow.net/questions/118052/symbols-of-elliptic-operators Comment by Branimir Ćaćić Branimir Ćaćić 2013-01-04T14:11:54Z 2013-01-04T14:11:54Z Do you allow for the various Clifford actions entering into $\mathscr{C}_n$ to come from differing inner products (viz, Riemannian metrics) on $\mathbb{R}^n$? http://mathoverflow.net/questions/117125/a-novice-question-on-quantum-mechanics/117133#117133 Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-24T10:55:26Z 2012-12-24T10:55:26Z Not just any matrices, though, but the positive (semi-definite) ones. http://mathoverflow.net/questions/116888/on-john-von-neumann-and-quantum-mechanics Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-21T17:44:41Z 2012-12-21T17:44:41Z You can probably strike the case of $L^\infty(M) \otimes F$ off your list of candidates, for almost-commutative spectral triples really are a &quot;semi-classical&quot; tool for obtaining <i>classical</i> field theories of a certain form. On the other hand, Connes--Rovelli's so-called &quot;thermal time hypothesis,&quot; which builds on Connes's work on type III factors, might be more germane? http://mathoverflow.net/questions/116860/spectrum-of-the-normal-operator-associated-to-compact-supported-spectral-measures Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-20T14:15:57Z 2012-12-20T14:15:57Z Yes, you're absolutely right. I imagine that the result you want is precisely the theorem on Page 7 of <a href="http://www.math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/RaskinSam.pdf" rel="nofollow">math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/&hellip;</a>, namely, that the support of a compactly-supported projection-valued measure is one and the same as the spectrum of the associated normal operator. http://mathoverflow.net/questions/116860/spectrum-of-the-normal-operator-associated-to-compact-supported-spectral-measures Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-20T12:32:52Z 2012-12-20T12:32:52Z By &quot;spectral&quot; do you mean a projection-valued measure? Because if so, then doesn't this hold just by definition? http://mathoverflow.net/questions/116133/explicit-computations-of-examples-in-spin-geometry Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-12T05:33:21Z 2012-12-12T05:33:21Z In terms of additional resources, pretty much all introductory accounts of spectral triples will at least sketch basic theory for and examples of Dirac operators. Joseph Varilly's lecture notes on spectral triples (<a href="http://toknotes.mimuw.edu.pl/sem3/index.html" rel="nofollow">toknotes.mimuw.edu.pl/sem3/index.html</a>) include very detailed, step-by-step exercises working through the spin geometry of the circle, 2-torus, and 2-sphere, though without covering any index theory at all. http://mathoverflow.net/questions/115309/examples-of-clifford-modules Comment by Branimir Ćaćić Branimir Ćaćić 2012-12-04T04:08:04Z 2012-12-04T04:08:04Z Those three examples, together with their twisted versions, really seem to be the main examples--poking about Berline, Getzler, and Vergne's &quot;Heat kernals and Dirac operators,&quot; the other main reference, brought up nothing more. For what it's worth, perhaps the most general way Clifford modules arise in the wild, at least when $M$ is even-dimensional, comes from seeking &quot;square roots&quot; to generalised Laplacians: $E \to M$ admits a Clifford action and compatible connection iff if it admits a Dirac-type operator, i.e., a first order differential operator $D$ such that $D^2$ is Laplace-type. http://mathoverflow.net/questions/114195/integral-kernel-for-the-resolvent-of-the-laplace-operator/114202#114202 Comment by Branimir Ćaćić Branimir Ćaćić 2012-11-24T03:35:41Z 2012-11-24T03:35:41Z Not complete nonsense, but certainly requires more care than what you outlined. On the other hand, if you should ever need to carry out these calculations in the curved case...