Hot answers tagged

3

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 ...


1

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible