Michael Murray
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Registered User
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Apr 27 |
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Automorphism of a Lie group which preserves a maximal torus is necessarily an inner automorphism? Assume $G$ compact. If your result was true then every automorphism would be inner. Indeed if $\mu$ was an automorphism you can find $g$ such that $Ad_g \mu (T) = T$. But $Aut(G)/Ad(G)$ is the non-trivial group of automorphisms of the Dynkin diagram. |
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Apr 5 |
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A k-form is thought of as measuring the flux through an infinitesimal k-parallelepiped Can you get hold of a copy of Misner, Thorne and Wheeler's Gravity ? They spend a lot of time explaining this point of view with great pictures. My memory (lost my copy in a postdoctoral move somewhere) is that what is measured is the flux of the $k$-form. |
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Mar 19 |
awarded | ● Yearling |
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Mar 15 |
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Fluid mechanics and topology Have a look for fluid mixing and know theory. Something like rsta.royalsocietypublishing.org/content/364/1849/… |
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Feb 21 |
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What is the exterior derivative intuitively? So the derivative of $\omega \colon X \to \Omega^p(X)$ at $x \in X$ is I guess the tangent map $T_x(\omega) \colon T_x X \to T_{\omega(x)} \Omega^p(X)$. How do you get the $p+1$-form? |
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Feb 12 |
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Pseudo-Differentialforms @Nevermind. Remember the co-ordinate change formula for the integral ? What comes out is the absolute value of the determinant of the jacobian of the co-ordinate change function. If you don't have an orientation this is how things have to transform to get a sensible definition of integral. |
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Feb 11 |
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Why does the group act on the right on the principal bundle? @Ben Webster. It gets even more fun if start worrying about identifying the Lie algebra on the group with left or right invariant vector fields and messing with the definition of connection on your left principal bundle ... |
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Feb 8 |
answered | Tools for long-distance collaboration |
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Feb 8 |
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group of diffeomorphisms of a manifold There is an explanation of the Frechet manifold structure on the group of diffeomorphisms of a manifold in Hamilton's paper: projecteuclid.org/… |
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Feb 8 |
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From Topological to Smooth and Holomorphic Vector Bundles @Daniel, @Ricardo. Thanks for the reminders that $\mathbb R$ has lots of different (although diffeomorphic) differentiable structures. Something I had forgotten! |
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Feb 8 |
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From Topological to Smooth and Holomorphic Vector Bundles Ricardo. Yes that was the question. Thanks. You could just call them equal couldn't you in such a case ? |
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Feb 8 |
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From Topological to Smooth and Holomorphic Vector Bundles What's your definition of equivalent Ricardo ? |
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Feb 8 |
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From Topological to Smooth and Holomorphic Vector Bundles I realise on reading the other responses that I may have misunderstood C. By de Rham cohomology of $E$ did you mean the de Rham cohomology of $E$ as a manifold or some sort of $E$ valued forms on $B$ cohomology ? |
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Feb 8 |
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From Topological to Smooth and Holomorphic Vector Bundles Thanks Daniel. I was little nervous about that point. I was hoping that because $B$ was fixed I could get away with a finite-dimensional Grassmanian but was't sure if it might change as I varied the maps homotopically? I think though the choice o finite dimensional Grassmanian can be fixed by the dimension of $B$? |
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Feb 7 |
answered | From Topological to Smooth and Holomorphic Vector Bundles |
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Feb 7 |
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From Topological to Smooth and Holomorphic Vector Bundles Siqi He: In your example there is no reason to expect they are vector bundles. |
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Jan 25 |
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Lie groups bundle An example to think about would be $U(2)/T \to U(2)/N_{U(2)}(T)$ which is $S^2 \to \mathbb{R}P_2$. The Weyl group is $\mathbb{Z}_2$ and $-1$ acts by the antipodal map on $S^2$. |
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Jan 25 |
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Lie groups bundle Note too Oscar that in your case if $H=T$ then $K/H$ is a (finite) group, the Weyl group of $G$ and your fibre bundle is a principal bundle for the Weyl group. |
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Jan 25 |
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Lie groups bundle Once you understand the associated bundle construction all that is missing is checking the map I have described is a fibre bundle map which is straightforward. |
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Jan 24 |
answered | Lie groups bundle |
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Jan 13 |
answered | Which popular games are the most mathematical? |
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Jan 10 |
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Eigenvalues of the products of a fixed unitari matrix with diagonal unitari matrices How are you ordering your eigenvalues ? |
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Jan 8 |
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Bifunctor: Vector space of linear transformations betw vector spaces as bifunctor How is $E$ a category? Shouldn't you be thinking of $E$ as an object in the category of vector spaces and linear maps. |
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Jan 6 |
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Should science authors discourage / boycott the recent push for author IDs Does anyone know if AMS has an opinion on ORCID? |
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Dec 31 |
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Is there formula name and proof for this theorem ? Sorry I deleted my earlier comment as it seemed irrelevant after you got the answer on MSE. Thanks for your reply. |
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Dec 31 |
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Is there formula name and proof for this theorem ? Your cross post over here has been answered. math.stackexchange.com/questions/268129/… |
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Dec 30 |
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trigonometric non-identity Wolframalpha says that $\sin((3+4π)/52)=0.2949023 \dots $ Does that help :-) |
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Dec 28 |
answered | n-categorical description of Chern classes |
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Dec 28 |
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Why is the Leibniz rule a definition for derivations? added 5 characters in body |
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Dec 28 |
answered | Why is the Leibniz rule a definition for derivations? |
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Dec 28 |
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Why is the Leibniz rule a definition for derivations? +1 for using tangency classes of paths instead of derivation as the definition of tangent vectors. Personally I think it makes calculations easier but that's a subjective judgement not a mathematical one. |
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Dec 24 |
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A novice question on Quantum Mechanics You can't add trace one matrices and get a trace one matrix though. I guess you are adding and renormalising ? Forming a convex combination as Konrad points out. |
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Dec 24 |
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A novice question on Quantum Mechanics But if states are rays in a Hilbert space then you have the problem that rays cannot be added. I think that is Ryan's concern. |
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Dec 24 |
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A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ Conjugation won't be transitive either. You can make $Isom(E_x, E_x)$ and $Isom(E_y, E_y)$ act freely and transitively on the left and right of $Isom(E_x, E_y)$, and they are groups isomorphic to $GL(n, \mathbb{R})$ but there is no canonical isomorphism. |
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Dec 24 |
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A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ But to be a principal bundle you need the action to be defined without choosing a trivialisation surely? In any case conjugation won't be a free action. |
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Dec 24 |
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A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ added 27 characters in body; deleted 60 characters in body; edited body |
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Dec 24 |
answered | A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ |
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Dec 23 |
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Quotient of a compact Lie group by maximal Torus Ah nice. Use the large cell in the Bruhat decomposition which is dense. |
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Dec 23 |
awarded | ● Enthusiast |
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Dec 22 |
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Quotient of a compact Lie group by maximal Torus Sorry I missed the word dense and was thinking discrete lattice! Your latitude and longitude idea should work. There might be a general approach using a dense grid in the Lie algebra mapped to the group with the exponential map. |
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Dec 22 |
revised |
Quotient of a compact Lie group by maximal Torus deleted 26 characters in body |
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Dec 22 |
answered | Quotient of a compact Lie group by maximal Torus |
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Dec 11 |
answered | Natural connection on U(1) principal bundles over S^2 with Chern number>1 |
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Nov 30 |
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Topology of the Universal Spinor Field Bundle I've added a link for the Hamilton paper. I'd be interested in a reference for the push-down result. |
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Nov 30 |
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Topology of the Universal Spinor Field Bundle Added link to reference. |
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Nov 26 |
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Topology of the Universal Spinor Field Bundle Forgot the reference. |
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Nov 26 |
answered | Topology of the Universal Spinor Field Bundle |
