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If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ... 2 (1)$M,\ N$has charts$x,\ y$Then if$\omega$is a pull back then$\omega= (f\circ \pi) (x,y) dy_{1}\cdots dy_p$where$f: N\rightarrow {\bf R}$(For convenience, we can write it) So $$\partial_{x_i} (f\circ \pi )= df\ d\pi \partial_{x_i} =0$$ If$d\pi X=0$, then$X=g_i(x,y)\partial_{x_i}$So$i_X\omega =0$In further since$L_X=i_Xd +di_X$so that ... 2 This post just consolidates the comments. If$V$is a real vector bundle, a fiberwise inner product on$V$determines an isomorphism$V \to V^*$by$v \mapsto \langle v, -\rangle$. So anything else you could possibly do to$V$and$V^*$are isomorphic: in particular, it determines an isomorphism$\Lambda^i T^*M \otimes V \to \Lambda^i T^*M \otimes V^*$. ... 1 Suppose that$V$is$n$-dimensional, the space of$n$-alternating forms defined on on$V$has dimension 1 and$det$is a generator. Let$f(A)=det(A^*)A$is multilinear and alternating. Thus there exists$c$such that$det(A^*)=cdet(A)$. In particular, if$A=I_n$the matrix of the identity map,$I^*=I$thus$det(I^*)=cdet(I)$implies$c=1$and for every ... 1 Hint This follows almost immediately from the coordinate-free characterization of$\det$: The map$T$determines a canonical map $$\det T : \Lambda^n V^* \to \Lambda^n V^*$$ defined by extending the map$v_1 \wedge \cdots \wedge v_n \mapsto T(v_1) \wedge \cdots \wedge T(v_n)$by linearity; here$n := \dim V$. (The existence of this map---and its ... 1 OK, I now have most of an answer, which I will post here. That said... any references would really be appreciated. I am still looking for references on the matter; I don't want to have to rederive this theory myself. Anyway, yes, it seems to be an alternate characterization of a$\lambda$-ring, via the usual way of converting between$e_n$(elementary ... 1 Your notation may be somewhat pratical in very low-dimensional spaces. But it has several strong drawbacks : Most mathematicians don't work in a fixed dimension with explicit indices, but in general dimension$n$with variable indices. So this would mean writing basis vectors as$e_{2^i}$instead of$e_i$, and$e_{2^i + 2^j+2^k}$instead of$e_{ijk}$for ... 1 First, it's pretty clear that the field$K$does not matter in any way (if you know how to prove things for$\mathbb{R}$, then just check that you don't use any special property of this field). Then, be careful : the statement for tensor algebras is already false for finite-dimensional vector spaces : if$V$has finite dimension,$T(V^*)$has countable ... 1 Try starting from the other direction and use$(F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta ...
Given an arbitrary $k$-element ordered list $i_1,i_2,\dots,i_k$ (consisting of distinct elements), choose complementary numbers $j_{k+1},\dots,j_n$ so that $\{i_1,\dots,i_k,j_{k+1},\dots,j_n\} = \{1,2,\dots,n\}$. If you order the $j$'s so that e_{i_1}\wedge e_{i_2} \wedge \dots \wedge e_{i_k}\wedge e_{j_{k+1}}\wedge \dots \wedge e_{j_n} = e_1\wedge ...