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1

If $E \to X$ is a complex vector bundle with structure group $G=U(1)\times SU(n-1) \subset U(n)$ there exists complex line bundle $E_1$ (with structure group $U(1)$) and a complex vector bundle $E_2$ with structure group $SU(n-1)$ such that $E$ is isomorphic to $E_1 \oplus E_2$. This makes the Chern classes of $E$ computable in the terms of chern classes of $...


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first of all if you have maps $f:X\to S$,$g:Y\to S$,by a $S$ morphism between $h:X\to Y$ we mean that $f=h\circ g$(so the two way you can go from $X$ to $S$ should be the Same). In your example you have maps $E_{|U}\to U$,$U\times V\to U$ so you can talk about a $U$ morphism between $E_{|U}$ and $U\times V$(the point of a $S$ morphism $h$ is basically that ...


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(Edited after some discussion in the comments.) As pointed out by Jonas, the four properties listed on Wikipedia do not characterise the Euler class as setting $e(E) = 0$ for all $E$ also satisfies those properties. I just want to point out that the property which Wikipedia calls normalization is redundant: it follows from the Whitney sum formula and the ...


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The standard reference for characteristic classes is the book by Milnor and Stasheff "Characteristic Classes", and I recommend reading it. Your second question is also answerd in these seminar notes ( Theorem 3.2) by Matthias Görg. The answer to your first question is no. They don't make sure we just have $e(E)=0$ for all bundles. Add a fifth axiom,...


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I still don't know how to solve the problems above, but now I have another approach to proof lemma 14.8,here I use Tietze extension theorem to gain the extension of isomorphism instead of using parallel transport. First, consider the manifold $(0,1) \times M$ and replaces the almost complex structure on $(1-\delta,1)$ and $(0,\delta)$ to be $j_1$ and $j_0$. ...


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The Thom-Isomorphism has a degree shift, if you use the natural $\mathbb{Z}$ grading on $K^*$, instead of the $\mathbb{Z}/2\mathbb{Z}$ grading. For your next questions, let $f:X \to Y$ be an embedding of oriented smooth manifolds. So we have $TX \overset{\sim}{=} f^*TY \oplus N(f)$, where $N(f) \to X$ is the normal bundle of $f$. Using the multiplicativity ...


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This is false. As an example, consider the circle $S^1$. All of the jet spaces of $C^\infty(S^1,\mathbb{R})$ are trivial, and since smooth functors on $\mathsf{Vect}_{\mathbb{R}}$ preserves identity morphisms, any bundle arising from these functors must also be trivial. This means that nontrivial bundles such as the Möbius strip cannot arise this way. Your ...


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$m = 1$. In fact, $$Q^\vee(1) \cong \wedge^{\mathrm{rank}(Q)-1}Q$$ and a wedge power of a globally generated vector bundle is globally generated. For isotropic Grassmannian the same is true by restriction.


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(There are many different vector bundles on a given space, so I'm not sure if by "the vector bundle" you really mean the tangent bundle. In any case, the tangent bundle is an example of a vector bundle that doesn't fit your definition.) Informally, the tangent bundle is the space of vectors "tangent to" or "inside" a manifold. ...


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According to the Connection_(vector_bundle) article of Wikipedia, connecions are $R$-linear map from $\Gamma(E)$ to $\Gamma (T^\ast M \otimes E)$ (both are $C^\infty(M)$-modules), but not $C^\infty(M)$-linear map since: $$\nabla(fσ)=f\nabla σ+df\otimes σ$$ Here $f \in C^\infty(M)$, the second term make it not $C^\infty (M)$-linear. But the subtraction of two ...


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Let $R$ be a ring (need not be commutative) and let $M$ and $N$ be left $R$-modules. As usual set $M^* = \hom_R(M,R)$ with its canonical right $R$-module structure. There is a natural map $$\Psi_{M,N} : M^* \otimes_R N \longrightarrow \hom_R(M,N)$$ such that $\Psi(f\otimes x)(y) = f(y)x$. Lemma. Fixing $M$, this map is an isomorphism for all $N$ if and only ...


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This is a special case of a general fact from EGA. Let $f:X\to S$ be a projective morphism and $F$ a coherent sheaf on $X$ flat over $S$. Then, there exists a complex of vector bundles $0\to E_0\to E_1\to\cdots$ which computes the direct images after any base change of $S$. Further, if $n$ is the maximum of the relative dimensions of all fibers, then we may ...


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It is more convenient to think here of $w\mathbb{P}$ as of a quotient stack; anyway, if $Y$ i smooth it does not pass through the stacky points $$ (0,0,0,1,0), (0,0,0,0,1) \in w\mathbb{P} $$ and therefore the stacky structure of $w\mathbb{P}$ plays no role for $Y$. The advantage of $w\mathbb{P} = \mathbb{P}(w_0,w_1,\dots,w_n)$ as of a stack is that it comes ...


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Yes, in general if $E$ is a holomorphic vector bundle over a Stein manifold $X$, then $H^p(X,\mathcal{O}(E))=0$ for all $p\geq 1$, by Cartan's theorem B.


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The correspondence is $\Phi\colon \Omega_{\rm hor}^q(P;V)^G \to \Omega^q(M;P\times_{\rho}V)$, given by $$(\Phi\omega)_x(v_1,\ldots, v_q) = [p, \omega_x(v_1^\uparrow,\ldots, v_q^\uparrow)]\tag{$\ast$},$$where $p \in P_x$ and the $v_k^\uparrow \in T_pP$ are lifts of $v_k$, i.e., satisfying ${\rm d}\pi_p(v_k^\uparrow)=v_k$. One has to prove that the expression ...


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