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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 $... 1 first of all if you have maps$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$is basically that ... 3 (Edited after some discussion in the comments.) As pointed out by Jonas, the four properties listed on Wikipedia do not characterise the Euler class as setting$e(E) = 0$for all$E$also satisfies those properties. I just want to point out that the property which Wikipedia calls normalization is redundant: it follows from the Whitney sum formula and the ... 4 The standard reference for characteristic classes is the book by Milnor and Stasheff "Characteristic Classes", and I recommend reading it. Your second question is also answerd in these seminar notes ( Theorem 3.2) by Matthias Görg. The answer to your first question is no. They don't make sure we just have$e(E)=0$for all bundles. Add a fifth axiom,... 0 I still don't know how to solve the problems above, but now I have another approach to proof lemma 14.8，here I use Tietze extension theorem to gain the extension of isomorphism instead of using parallel transport. First, consider the manifold$(0,1) \times M$and replaces the almost complex structure on$(1-\delta,1)$and$(0,\delta)$to be$j_1$and$j_0$. ... 1 The Thom-Isomorphism has a degree shift, if you use the natural$\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$. Using the multiplicativity ... 0 This is false. As an example, consider the circle$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}}$preserves identity morphisms, any bundle arising from these functors must also be trivial. This means that nontrivial bundles such as the Möbius strip cannot arise this way. Your ... 0$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. 3 (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. ... 0 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 ... 4 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 ... 0 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 ... 2 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 ... 2 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. 1 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 ...