User michael murray - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:37:37Z http://mathoverflow.net/feeds/user/13762 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49384/tools-for-long-distance-collaboration/121184#121184 Answer by Michael Murray for Tools for long-distance collaboration Michael Murray 2013-02-08T12:43:06Z 2013-02-08T12:43:06Z <p>Has anyone tried one of these Logitech conference cameras with Skype </p> <p><a href="http://www.logitech.com/en-us/product/Conferencecam?crid=1252" rel="nofollow">http://www.logitech.com/en-us/product/Conferencecam?crid=1252</a></p> <p>as a collaborative tool. I saw one report on-line of someone trying to use it to look at a whiteboard and claiming it didn't work because of reflections of the shiny surface.</p> <p>Thanks - Michael</p> http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121063#121063 Answer by Michael Murray for From Topological to Smooth and Holomorphic Vector Bundles Michael Murray 2013-02-07T12:46:39Z 2013-02-07T12:46:39Z <p>The answer to A is no. The topological bundle is determined by a continuous homotopy class of maps into the classifying space. Choosing a compatible smooth structure means picking a smooth map in that class. Any two such choices are smoothly homotopic. B however is true. Look up the Jacobian of a Riemann surface. How are you going to grade in C ? Where are the forms on $E$ of degree higher then the dimension of $B$?</p> http://mathoverflow.net/questions/119751/lie-groups-bundle/119760#119760 Answer by Michael Murray for Lie groups bundle Michael Murray 2013-01-24T13:54:29Z 2013-01-24T13:54:29Z <p>If you are prepared to accept that $G \to G/K$ is a principal $K$-bundle there is an easy proof. You have that $K$ acts on the homogeneous space $K/H$ so you have an associated fibre bundle $$\frac{G \times K/H}{K} \to G/K$$ with fibre $K/H$. The total space is actually $G/H$. You can construct a fibre bundle isomorphism from this to $G/H$ by $$gH \mapsto [g, H]_K$$ where $[g, H]_K$ is the orbit under the $K$ action $(g, H)k = (gk, k^{-1}H)$. Why is this well defined ? You can check that $$ghH \mapsto [gh, H]_K = [g, hH]_K = [g, H]_K$$ as $H \subset K$. It has an inverse which is $$[g, kH]_K \mapsto gkH$$ This is also well-defined as $$[gk_1, k_1^{-1}kH]_K \mapsto gk_1 k_1^{-1} k H = gk H$$</p> http://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical/118807#118807 Answer by Michael Murray for Which popular games are the most mathematical? Michael Murray 2013-01-13T13:09:34Z 2013-01-13T13:09:34Z <p><a href="http://en.wikipedia.org/wiki/Rummikub" rel="nofollow">Rummikub</a> ? It encourages some logical thought and analysis. It seems to have at least one mathematical paper on it </p> <p><a href="http://comjnl.oxfordjournals.org/content/49/6/665.abstract" rel="nofollow">http://comjnl.oxfordjournals.org/content/49/6/665.abstract</a></p> <p>and it's popular and fun. </p> http://mathoverflow.net/questions/117378/n-categorical-description-of-chern-classes/117407#117407 Answer by Michael Murray for n-categorical description of Chern classes Michael Murray 2012-12-28T13:51:47Z 2012-12-28T13:51:47Z <p>The answer will depend on your realisation of a $k$-circle bundle. In the case of $i=2$ (second Chern class) there are results associating to any principal $G$-bundle a bundle $2$-gerbe. See: </p> <p>Bundle Gerbes for Chern-Simons and Wess-Zumino-Witten Theories. Alan L. Carey, Stuart Johnson, Michael K. Murray, Danny Stevenson and Bai-Ling Wang. Communications in Mathematical Physics, 159 (3) (2005), 577-613 math.DG/0410013</p> <p>and the references there in to Danny Stevenson and Stuart Johnson's PhD theses and papers. Of course you have to be happy that a $3$-circle bundle is a $2$-gerbe.</p> <p>More generally you might find something useful in:</p> <p>P. Gajer Geometry of Deligne cohomology, Invent. Math. 127 (1997), 155-207.</p> <p>which gives a realisation of principal $B^k \mathbb{C}^*$ bundles which are another possible way of realising $(k+1)$-circle bundles or at least mathematical objects determined by a characteristic class in degree $H^{k+1}(M, \mathbb{Z})$. There is a nice inductive classifying theory and a simplicial realisation of these spaces.</p> http://mathoverflow.net/questions/117374/why-is-the-leibniz-rule-a-definition-for-derivations/117385#117385 Answer by Michael Murray for Why is the Leibniz rule a definition for derivations? Michael Murray 2012-12-28T04:26:46Z 2012-12-28T04:37:57Z <p>If this is your first time doing differential geometry you should do the calculation that ayanta is referring to in their comment. Let $X \colon C^\infty(M) \to \mathbb{R}$ be a linear map satisfying the Leibniz rule $$X(fg) = X(f) g(x) + f(x) X(g)$$ for all $f, g, \in C^\infty(M)$. Show that if $\psi = (\psi^1, \dots, \psi^n)$ are co-ordinates in any neighbourhood of $x$ then there are real numbers $X^1, \dots, X^n$ such that for all $f \in C^\infty(M)$ we have $$X(f) = \sum_{i=1}^n X^i \frac{\partial (f \circ \psi^{-1})}{\partial \psi^i} (\psi(x)) .$$</p> <p>Also a good exercise to show that for the same $X$ there is a smooth function $\gamma \colon (-\epsilon, \epsilon) \to M$ such that $\gamma(0) = x$ and for any $f \in C^\infty(M)$ we have $$X(f) = \frac{ d (f \circ \gamma) }{ dt}(0).$$ This connects you to tangent vectors thought of as equivalence classes of curves as Davidac897 discusses.</p> http://mathoverflow.net/questions/117075/a-question-on-isomp-1e-p-2e-rightrightarrows-x/117120#117120 Answer by Michael Murray for A question on $Isom(p_1^*E,p_2^*E) \rightrightarrows X$ Michael Murray 2012-12-24T03:36:08Z 2012-12-24T03:41:44Z <p>I'm not an algebraic geometer so let's call $X$ a real manifold. I don't think that really matters it could be a topological space or even a set for what I am about to say. Assume that $E \to X$ is a rank $n$ real vector bundle. For any $x \in X$ let $E_x$ be the fibre of $E$ over $x$. The natural structure you have in this situation is that if $f \colon E_{x} \to E_{y}$ and $g \colon E_{y} \to E_{z}$ are isomorphisms then you can compose to get $g \circ f \colon E_{x} \to E_{z}$ also an isomorphism. From this follows the fact that you have a groupoid whose objects are all $x \in X$ and whose morphisms from $x$ to $y$ are $Isom(E_x, E_y)$ (or $Isom(E_y, E_x)$ depending on how you like to compose morphisms.)</p> <p>(1) $Isom(E, E)$ is a perfectly reasonable object, it's a bundle of groups over $X$. But it doesn't capture all the information such as isomorphisms from $E_x$ to $E_y$ where $x \neq y$. </p> <p>(2) $Isom(p_1^*E, p_2^*E)$ is the union over all $x, y \in X$ of $Isom(E_x, E_y)$ and if $f \in Isom(E_x, E_y)$ then the two maps are $f \mapsto x$ and $f \mapsto y)$. </p> <p>(3) I am not sure of the answer to this but it seems reasonable to me that this groupoid captures information about the symmetries of $E \to X$.</p> <p>I don't think that $Isom(p_1^*E, p_2^*E)$ being a principal $GL(n, \mathbb{R})$ bundle is correct. I don't see any reason why $Isom(E_x, E_y)$, for example, is acted on by $GL(n, \mathbb{R})$. </p> http://mathoverflow.net/questions/117019/quotient-of-a-compact-lie-group-by-maximal-torus/117025#117025 Answer by Michael Murray for Quotient of a compact Lie group by maximal Torus Michael Murray 2012-12-22T12:03:08Z 2012-12-22T14:31:32Z <p>Maybe try first with $G=SL(2, \mathbb{C})$, $K = SU(2)$ and $T = U(1)$, the diagonal matrices with determinant $1$. Then $K/T = S^2$. </p> <p>Edit: I was thinking of a discrete lattice so this answer which I thought was a counter example isn't.</p> http://mathoverflow.net/questions/116074/natural-connection-on-u1-principal-bundles-over-s2-with-chern-number1/116075#116075 Answer by Michael Murray for Natural connection on U(1) principal bundles over S^2 with Chern number>1 Michael Murray 2012-12-11T12:32:17Z 2012-12-11T12:32:17Z <p>If you work out how the circle group acts on the fibres then the connection form has to pull-back to the Maurer-Cartan form on the circle group. That will fix the multiple. </p> http://mathoverflow.net/questions/114430/topology-of-the-universal-spinor-field-bundle/114500#114500 Answer by Michael Murray for Topology of the Universal Spinor Field Bundle Michael Murray 2012-11-26T08:23:16Z 2012-11-30T01:57:10Z <p>If you are happy with Frechet bundles here is an alternative approach. Let $\pi \colon {\mathcal G} \times M \to M$ be the projection and consider $\pi^{-1}(TM) \to {\mathcal G} \times M$ a real vector bundle of rank $n$. Following the notation in the paper let $P_{GL^+} \to M$ be the $GL^+(n, {\mathbb R})$ bundle of oriented frames of $TM$. The bundle of oriented frames of $\pi^{-1}(TM)$ is $\pi^{-1}(P_{GL^+})$. As in the paper pick a lift $P_{\widetilde{GL}^+} \to M$ of $P_{GL^+}$ to $\widetilde{GL}^+(n, {\mathbb R}) \to M$ of bundles over $M$. This exists because we assume $M$ is spin. Then $\pi^{-1}(P_{\widetilde{GL}^+})$ is a lift of $\pi^{-1}(P_{GL^+})$ to $\widetilde{GL}^+(n, {\mathbb R})$. </p> <p>If $g \in {\mathcal G}$ and $m \in M$ then $\pi^{-1}(TM)_{(g, m)} = T_m M$ so has on it an inner product defined by $g(m)$. Denote this "universal" inner product on $\pi^{-1}(TM)$ by $g$. It will be smooth for the usual reason with Frechet manifolds which is because if $M$ and $N$ are finite-dimensional bundles then the evaluation map $$M \times C^\infty(M, N) \to N$$ is a smooth map of Frechet manifolds [1]. Again following the approach in the paper we let $P_{SO} \subset \pi^{-1}(P_{GL^+})$ be the subbundle of oriented orthonormal frames for the metric $g$. Taking the pre-image of this in $\pi^{-1}(P_{\widetilde{GL}^+})$ gives us a $Spin(r, s)$ bundle over $\mathcal{G} \times M$. The associated vector bundle to this using the spin representation gives us $E$ as a smooth, finite rank, Frechet vector bundle. </p> <p>Finally you want a theorem that says that when you "push-down" $E$ with $\pi$ the result is a smooth Frechet vector bundle on ${\mathcal G}$. This seems reasonably but I'm not sure where to find it. I can't see it in [1].</p> <p>Sorry this is a bit sketchy but that reflects the sketchiness of my knowledge of Frechet manifolds. </p> <p>[1] Richard Hamilton -- The Inverse Function Theorem of Nash Moser. <a href="http://dx.doi.org/10.1090%2FS0273-0979-1982-15004-2" rel="nofollow">http://dx.doi.org/10.1090%2FS0273-0979-1982-15004-2</a></p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112066#112066 Answer by Michael Murray for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? Michael Murray 2012-11-11T08:09:13Z 2012-11-11T12:24:42Z <p>Why are you not happy with using Grassmanians as in:</p> <p><a href="http://en.wikipedia.org/wiki/Classifying_space_for_U(n" rel="nofollow">http://en.wikipedia.org/wiki/Classifying_space_for_U(n</a>) ?</p> <p>A related approach that might interest you is given in Dupont's book</p> <p><a href="http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633" rel="nofollow">http://www.amazon.com/Curvature-Characteristic-Classes-Lecture-Mathematics/dp/3540086633</a></p> <p>using simplicial manifolds. A simplicial manifold $X$ is a sequence of manifolds $\lbrace X_n \rbrace$ and various maps between them. From a simplicial manifold you can construct a topological space called its realisation. This is how you define $EG \to BG$. Although it isn't a manifold you can realise its topology using the finite dimensional spaces $X_n$ which is how you tie things back to the statement of Chern-Weil theory given in the question. In this example the simplicial space is also the one arising in the bar construction and Milnor's join construction of $EG \to BG$.</p> http://mathoverflow.net/questions/111430/tangent-bundle-and-normal-bundle-in-self-product/111431#111431 Answer by Michael Murray for Tangent bundle and normal bundle in self-product Michael Murray 2012-11-04T06:05:08Z 2012-11-04T07:36:50Z <p>In the case of differential geometry everything reduces to vector spaces. Let $x \in X$. Then at any point $(x, y) \in X \times X$ $$T_{(x, y)} X \times X = T_x X \oplus T_y X .$$ Using this identification the tangent to the diagonal at a point $(x, x)$ is the subspace of $T_{(x, x)} X \times X$ given by $$T_{(x, x)} (\Delta) = \lbrace (\xi, \xi ) \mid \xi \in T_x X \rbrace \subset T_{(x, x)} X \times X.$$ On the other hand the normal is the quotient $$(T_{(x, x)} X \times X) / T_{(x, x)} \Delta$$ and we can identify this with $T_x X$ in at least two slightly different ways. Either $$\iota_1 \colon \xi \mapsto (\xi , - \xi) + T_{(x, x)} \Delta$$ or $$\iota_2 \colon \xi \mapsto (-\xi , \xi) + T_{(x, x)} \Delta .$$</p> <p>We can of course also identify $T_x X$ and $T_{(x, x)} \Delta$ by $\xi \mapsto (\xi, \xi)$. </p> <p>I guess one explanation for the two identifications of the normal bundle is that there is involution $\tau \colon X \times X \to X \times X$ given by $\tau(x, y) = (y, x)$ which fixes the diagonal pointwise and hence acts trivially on the tangent space to the diagonal. As a result it descends to an action on the normal bundle which interchanges the two identifications $\iota_1$ and $\iota_2$, that is $\tau \circ \iota_1 = \iota_2$</p> http://mathoverflow.net/questions/110797/a-sudden-smiley/110798#110798 Answer by Michael Murray for A sudden smiley? :-) Michael Murray 2012-10-27T00:59:27Z 2012-10-27T00:59:27Z <p>Surely you can draw the 2-D image in the XY plane so it consists of points of the form (x, y, 0) and then give each point in it a random non-zero Z co-ordinate. So it should look like a mess except viewed looking in along the Z-axis. </p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/110416#110416 Answer by Michael Murray for Slick ways to make annoying verifications Michael Murray 2012-10-23T10:56:11Z 2012-10-23T23:40:31Z <p>Elementary but still useful is the regular value theorem or the submersion theorem:</p> <p>Let $f \colon M \to N$ be smooth and $n \in N$. If $T(f)_m \colon T_m M \to T_{f(m)} N$ is onto for all $m \in f^{-1}(n)$ then $f^{-1}(n) \subset M$ is a submanifold of dimension $\text{dim}(N) - \text{dim}(M)$. </p> http://mathoverflow.net/questions/24526/good-papers-books-essays-about-the-thought-process-behind-mathematical-research/107008#107008 Answer by Michael Murray for Good papers/books/essays about the thought process behind mathematical research Michael Murray 2012-09-12T13:20:18Z 2012-09-12T13:20:18Z <p>Atiyah's <em>Advice to a Young Mathematician</em> is worth a look:</p> <p><a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf" rel="nofollow">http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf</a></p> <p>Michael</p> http://mathoverflow.net/questions/106623/vector-space-structure-on-velocity-space-of-manifold/106668#106668 Answer by Michael Murray for Vector space structure on velocity space of manifold Michael Murray 2012-09-08T12:55:27Z 2012-09-08T13:02:46Z <p>Nice answer BS. I was about to post something similar but I didn't have a proof of the non-linearity of the action of $k$-jets of diffeomorphisms. One additional remark that might help the OP is that you can put a vector space structure on $J^k_0({\mathbb R}^n, M)_x$ if you have additional structure on $M$. Basically you need to be able to determine a family of co-ordinates (actually $k$-jets of co-ordinates) that are related by linear transformations to avoid the non-linear action of diffeomorphisms when you change co-ordinates. Also you want to choose different co-ordinates at each point of $M$. Sufficient would be to choose at each $x \in M$ the $k$-jet of a diffeomorphism from $M$ to $T_xM$ sending $x$ to $0 \in T_xM$. For example if $M$ is Riemannian the $k$-jet of the inverse of the exponential map would do or if $M$ is a submanifold orthogonal projection onto the tangent subspace would work. In such a case composition with the chosen $k$-jet of a diffeomorphism defines a bijection $$J^k_0({\mathbb R}^n, M)_x \to J^k_0({\mathbb R}^n, T_xM)_0$$ and the latter space is a vector space because $T_xM$ is a vector space.</p> http://mathoverflow.net/questions/105217/how-to-easy-calculate-card-of-union-of-sets/105223#105223 Answer by Michael Murray for How to easy calculate card of union of sets? Michael Murray 2012-08-22T10:47:01Z 2012-08-23T17:38:55Z <p>To extend the last sentence of Ilya's reply have a look in Wikipedia for the <a href="http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle" rel="nofollow">inclusion-exclusion principle</a> which accounts for the size of the intersections by the formula</p> <p>$$\left| A \cup B \cup C \right | = \left| A \right| + \left| B \right| + \left| C \right| - \left| A \cap B\right| - \left| B \cap C \right| - \left| A \cap C \right| + \left| A \cap B \cap C\right|.$$</p> <p>There is an obvious generalisation to an alternating sum over the cardinalities of all the $k$-fold intersections in the case of $n$ sets. Of course if you don't know the cardinalities of the intersections this is not so useful!</p> http://mathoverflow.net/questions/104974/nontrivial-examples-of-non-trivial-principal-circle-bundles/105022#105022 Answer by Michael Murray for Nontrivial examples of non-trivial principal circle bundles Michael Murray 2012-08-19T05:22:40Z 2012-08-19T06:05:05Z <p>A third way to think about Anton Petrunin's example is that $S^2 \to {\mathbb R}P^2$ is a ${\mathbb Z}_2$ principal bundle where the action of ${\mathbb Z}_2 = \lbrace +1, -1 \rbrace$ is the obvious action on vectors in ${\mathbb R}^3$. As ${\mathbb Z}_2$ is a subgroup of $U(1)$, the circle group of complex numbers of length one, you can use standard principal bundle constructions to extend this ${\mathbb Z}_2$ principal bundle to a $U(1)$ principal bundle. In the case at hand these constructions just give the quotient of $S^2 \times S^1$ by ${\mathbb Z}_2$ as above. </p> <p>You can compute the transition functions of this $U(1)$ bundle explicitly with respect to the standard open cover of ${\mathbb R}P^2$ by just computing the same for the ${\mathbb Z}_2$ bundle $S^2 \to {\mathbb R}P^2$ and check that they give you a Cech representative for the non-zero class in $H^2({\mathbb R}P^2, {\mathbb Z}_2) = {\mathbb Z}_2$. </p> <p>The fact that this $U(1)$ bundle has a reduction to ${\mathbb Z}_2$ also tells us that when we square it we will get a trivial $U(1)$ bundle. Just think of squaring the ${\mathbb Z}_2$ valued transitions functions to get transitions functions for the squared bundle. Obviously they all just take the value $1$. </p> http://mathoverflow.net/questions/19930/writing-papers-in-pre-latex-era/102140#102140 Answer by Michael Murray for Writing papers in pre-LaTeX era? Michael Murray 2012-07-13T13:08:54Z 2012-07-13T13:08:54Z <p>There was, of course, a time when cut and paste meant cut and paste! I don't remember that but I do remember replacing sections of my thesis with sellotape. Another skill we have lost is trying to replace a small whited-out section of text with a word or two conveying the correct meaning but fitting in the gap!</p> <p><a href="http://www.maths.adelaide.edu.au/michael.murray/thesis.pdf" rel="nofollow">My thesis</a> from 1983 was typed by my supervisor's secretary on an IBM golfball with the occasional symbol written in. I didn't start using $\TeX$ until the first Apple LaserWriter. </p> http://mathoverflow.net/questions/128876/automorphism-of-a-lie-group-which-preserves-a-maximal-torus-is-necessarily-an-inn Comment by Michael Murray Michael Murray 2013-04-27T13:12:51Z 2013-04-27T13:12:51Z 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. http://mathoverflow.net/questions/126608/a-k-form-is-thought-of-as-measuring-the-flux-through-an-infinitesimal-k-parallele Comment by Michael Murray Michael Murray 2013-04-05T13:08:42Z 2013-04-05T13:08:42Z 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. http://mathoverflow.net/questions/124568/fluid-mechanics-and-topology Comment by Michael Murray Michael Murray 2013-03-15T02:34:29Z 2013-03-15T02:34:29Z Have a look for fluid mixing and know theory. Something like <a href="http://rsta.royalsocietypublishing.org/content/364/1849/3251" rel="nofollow">rsta.royalsocietypublishing.org/content/364/1849/&hellip;</a> http://mathoverflow.net/questions/21024/what-is-the-exterior-derivative-intuitively/56008#56008 Comment by Michael Murray Michael Murray 2013-02-21T09:55:07Z 2013-02-21T09:55:07Z 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? http://mathoverflow.net/questions/122268/where-do-mathematicians-go-to-discuss-research-grants-and-projects Comment by Michael Murray Michael Murray 2013-02-19T08:30:41Z 2013-02-19T08:30:41Z Has anyone seen &quot;increasing use of crowdfunding by scientists and mathematicians seeking research dollars&quot; ? http://mathoverflow.net/questions/121076/pseudo-differentialforms/121541#121541 Comment by Michael Murray Michael Murray 2013-02-12T11:09:01Z 2013-02-12T11:09:01Z @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. http://mathoverflow.net/questions/50473/why-does-the-group-act-on-the-right-on-the-principal-bundle Comment by Michael Murray Michael Murray 2013-02-11T07:43:02Z 2013-02-11T07:43:02Z @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 ... http://mathoverflow.net/questions/121168/group-of-diffeomorphisms-of-a-manifold Comment by Michael Murray Michael Murray 2013-02-08T10:38:30Z 2013-02-08T10:38:30Z There is an explanation of the Frechet manifold structure on the group of diffeomorphisms of a manifold in Hamilton's paper: <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183549049" rel="nofollow">projecteuclid.org/&hellip;</a> http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121150#121150 Comment by Michael Murray Michael Murray 2013-02-08T06:21:01Z 2013-02-08T06:21:01Z @Daniel, @Ricardo. Thanks for the reminders that $\mathbb R$ has lots of different (although diffeomorphic) differentiable structures. Something I had forgotten! http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121150#121150 Comment by Michael Murray Michael Murray 2013-02-08T06:17:07Z 2013-02-08T06:17:07Z Ricardo. Yes that was the question. Thanks. You could just call them equal couldn't you in such a case ? http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121150#121150 Comment by Michael Murray Michael Murray 2013-02-08T06:08:30Z 2013-02-08T06:08:30Z What's your definition of equivalent Ricardo ? http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121063#121063 Comment by Michael Murray Michael Murray 2013-02-08T02:14:12Z 2013-02-08T02:14:12Z 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 ? http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles/121098#121098 Comment by Michael Murray Michael Murray 2013-02-08T01:30:35Z 2013-02-08T01:30:35Z 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 $B$? http://mathoverflow.net/questions/121057/from-topological-to-smooth-and-holomorphic-vector-bundles Comment by Michael Murray Michael Murray 2013-02-07T12:30:25Z 2013-02-07T12:30:25Z Siqi He: In your example there is no reason to expect they are vector bundles. http://mathoverflow.net/questions/119751/lie-groups-bundle/119760#119760 Comment by Michael Murray Michael Murray 2013-01-25T09:23:58Z 2013-01-25T09:23:58Z 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 $S^2$.