# Olivier Bégassat

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olivier dot begassat dot cours at gmail dot com

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 2d comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$? Usually you'd include the 0-th exterior power $\Lambda^0(T^*E)$ (which is a line bundle) in the definition of $\Lambda(T^*E)$; in any case, $\Lambda(T^*E)$ (with the correct definition) is a vector bundle of dimension $2^n$ over $E$ (not $T^*E$) where, I take it, $E$ is a manifold of (real) dimension $n$. 2d comment Is $\Lambda(T^{*}E)=\bigoplus_{k=0}^n\Lambda^k(T^{*}E)$ a complex line bundle over $T^{*}E$? There seems to be a lot of confusion in your question. Are you sure that is what you mean to ask? As I understand it, the answer is trivially no. A vector bundle that decomposes as a direct sum of vector bundles has dimension equal to the sum of the dimensions of the vector bundles that appear in the sum... 2d answered Klein Bottle Embedding on $\mathbb{R}^4$. 2d awarded Nice Question Apr18 comment Klein Bottle Embedding on $\mathbb{R}^4$. It's not the antipode you are using. When he applies the antipode of $\Bbb R^3$ to a point of the torus of revolution, one of the circle coordinates, the "horizontal" one, if you see the tours as a vertical circle being rotated along the horizontal circle $C=\lbrace(x,y,z)\mid z=0\text{ and }x^2+y^2=1\rbrace$) is shifted by 180° (which corresponds to the $x\mapsto x+\frac12$) while the second circle coordinate is being "complex conjugated" (which corresponds to the $y\mapsto-y$). Apr18 comment Klein Bottle Embedding on $\mathbb{R}^4$. It is. I'm interested, how does Do Carmo define the Klein bottle? Apr18 comment Klein Bottle Embedding on $\mathbb{R}^4$. Sorry, the first map I wrote is wrong, I meant $(x,y)\mapsto(x+\frac12,-y)$. I think the one that is likely to work is $(x,y)\mapsto(x+\frac12,-y)$. Apr18 comment Klein Bottle Embedding on $\mathbb{R}^4$. What map are you talking about? Also, you should add @myname in front of your question if you want me to be notified. (Although I was still notified for some reason, how did that happen???) Apr18 comment Klein Bottle Embedding on $\mathbb{R}^4$. There's fundamentally a problem with $A$: it doesn't reverse orientation, so the quotient space associated to it is still orientable (whereas the Klein bottle isn't). You can try and see if $G$ is invariant under the map $(x,y)\mapsto (x,-y)$ or $(x,y)\mapsto (-x,y+\frac12)$. Apr18 comment Homology Whitehead theorem for non simply connected spaces That's very interesting. I asked John Klein for a reference in the comment section of his answer; do you know of a reference for this? Apr17 comment Is the tangent bundle of an oriented surface plus a trivial bundle trivial? You can view $E$ as the pullback of the (trivial) tangent bundle of $\Bbb R^3$ for some embedding $i:\Sigma\hookrightarrow \Bbb R^3$. You get $i^*(T_{\Bbb R^3})\simeq T_{\Sigma}\oplus\nu$, where $\nu$ is the normal bundle, and is the trivial line bundle by orientability of $\Sigma$. Hence $E$ is trivial. Apr17 comment How to determine the side on which a point lies? My bad, I got confused and for a moment ^^ Apr17 comment How to determine the side on which a point lies? This doesn't make much sense, you're not picking any "side" of your hyperplane this way. Apr17 comment Explain “There is a unique function from $\emptyset$ to any set $A$”. There is but one way to nothing at all. Apr17 comment Definition of General Associativity for binary operations I think it's a very good question, +1. You could certainly define an abstract bracketing as a rooted planar binary tree. Maybe this really overstating it, but "general associativity" appears as some kind of coherence theorem, not unlike the coherence theorem of Mac Lane. Apr16 comment Is C（E）a dual of any linear norm space? What do you mean by "$E$ is a closed bounded set"? Of what? Some $\Bbb R^n$? Or do you want $E$ to be any (presumably) compact Hausdorff space? In which case you'll also want to ask that $E$ be infinite, just to rule trivialties. Apr16 revised Proving that the fundamental group of Klein bottle is generated by two elements without using covering spaces and van Kampen theorem, etc. added 664 characters in body Apr16 answered Proving that the fundamental group of Klein bottle is generated by two elements without using covering spaces and van Kampen theorem, etc. Apr16 comment basic question of topology involving compactness and convexity As @dmw64 points out, you need to add a hypothesis to your question. Apr15 comment Show that if B is simply-connected, then p is a homeomorphism. It works, but you'll want to be precise and say that $\psi$ is homotopic rel. endpoints to a constant path, for any path is freely homotopic to a constant path.