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4

There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively. In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.


4

I don't know what kind of answer you're expecting at this level of generality. $\wedge^k V$ is irreducible as a representation of $GL(V)$, so in some sense there is no additional decomposition that can be done knowing nothing about the group $G$. Given the character of $V$, you can compute the character of $\wedge^k V$. For example, $$\chi_{\wedge^2 V}(g) ...


4

First of all, notice that exterior powers commute with base change (Eisenbud, Commutative Algebra with a View..., Proposition A2.2, p. 576), hence $$\Lambda^n_{F[x]} (F[x] \otimes_F V)=F[x] \otimes_F \Lambda^n_{F}V$$ You can easily check, that the following diagram (of $F$-modules) commutes ($m(\lambda)$ is the map $x \mapsto \lambda$ from the other ...


4

For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$. For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.) ...


3

They both may be right. If wedge product differs (Similar problem) and we set the definition of exterior derivative as $$d\omega=\sum_{I}d\omega_I\wedge dx^I,$$ then $d$ may differs as well (cause $\wedge$ appears). If we take axiomatic approach to exterior derivative, then one of axioms says ...


3

In general, consider a one form $\beta = \sum_i \beta_i dx^i $ on $U\subset\mathbb R^m$. If $$\| \beta\|^2 = \beta_1^2 + \cdots + \beta_m^2 \neq 0$$ on $U$, then the $(n-1)$-form $$\alpha = \frac{1}{\|\beta\|^2} \sum_i (-1)^{i-1} \beta_i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots \wedge dx^m$$ on $U$ safisfies $$\beta \wedge \alpha = dx^1 ...


3

The easiest way to answer your question is first bring it in an 'equivalent' form and then answer it. Of course you might still argue whether the the reversed form is really equivalent... The modified question is: Can we identify $e_1 \wedge \cdots \wedge \hat{e}_i \wedge \cdots \wedge e_N$ with $e_i^\perp$? (So I only moved the $\perp$ to the other side ...


3

This is not really a direct answer but is just too long for a comment. One curious fact about the wedge construction is that $\bigwedge^n V$ can be (functorially) realized either as a subspace of $\bigotimes^n V$ or as a quotient. (These realizations are canonically isomorphic when the characteristic of the underlying field is $0$ or greater than $n$, but ...


3

The Lie bracket is an alternating bilinear map. This means you can describe it in three different ways: As a function $\mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$ (which is alternating and bilinear), As a linear map $\mathfrak{g} \otimes \mathfrak{g} \to \mathfrak{g}$ (which is alternating), or As a linear map $\Lambda^2 \mathfrak{g} \to ...


3

Since the product is alternating, $$\color{red}{e_1}\wedge\color{blue}{e_1} = -\color{blue}{e_1}\wedge\color{red}{e_1}$$ and therefore $$e_1\wedge e_1 = 0$$


3

This is true if and only if the normal bundle to $\xi$ is trivial. In one direction, if $N(\xi) = TM/\xi$ is trivial, the bundle map $TM \to N(\xi) \to \Bbb R$ defines a 1-form $\alpha$ as desired (recalling that $T^*M = \text{Hom}(TM,\Bbb R)$ as vector bundles). On the other hand, given such an $\alpha$, put a Riemannian metric on $M$ and consider the ...


3

This follows from the fact that $\bigwedge^n (T^*M)$ (in other words, the space of $n$-forms on an $n$-dimensional manifold) is 1-dimensional. Since the determinant is one such form, all others are scalar multiples of it.


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

No, no, every $2$-vector can be written as the wedge product of two vectors. For example, assuming $c\ne 0$, you can write $$ae_1\wedge e_2+ be_1\wedge e_3+ce_2\wedge e_3 = (be_1+ce_2)\wedge (-\frac ace_1+e_3).$$


2

Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why $\det A=\det(E_1)\cdots\det(E_k)\det(D)\det(F_1)\cdots\det(F_\ell)$ and the ...


2

The $\mathbb{Z}_2$ grading is easy enough to anticipate. Given an integer it is either even or odd. So, there's your grading. A one-form is odd. A two-form is even. Even elements commute with all other elements under the wedge product whereas the product of odd elements anticommute. All of this is plainly seen in: $$ \alpha \wedge \beta = (-1)^{pq} \beta ...


2

That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed. However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare ...


2

I'll propose to you another (slightly different, but isomorphic) definition of the adjugate (classical adjoint). Im borrowing from section 8 of http://people.reed.edu/~jerry/332/27exterior.pdf . Let $f:V\rightarrow V$ (with $n$ the dimension of $V$). We have a canonical isomorphism $\phi:V=\wedge^1 V\rightarrow\mathrm{Hom}(\wedge^{n-1} V,\wedge^n V)$ ...


2

So let's assume that $V$ has a non-degenerate bilinear form $\langle\cdot,\cdot\rangle$ with a basis $e_1,\dots,e_n$ such that $\langle e_i,e_j\rangle = \delta_{ij}$, the Kronecker delta. Let $*$ denote the Hodge star operator. Note that we have the formula $$ \langle x,y\rangle = *((*x)\wedge y) .$$ Let's identify any operator on $V$ with its matrix ...


2

Comments to the question (v2): One can define left and right exterior derivatives $$ d_L(\omega\wedge\eta)~=~(d_L\omega)\wedge\eta + (-1)^{|\omega|}\omega\wedge d_L\eta \tag{L}, $$ $$ d_R(\omega\wedge\eta)~=~(-1)^{|\eta|} (d_R\omega)\wedge\eta + \omega\wedge d_R\eta \tag{R}, $$ $$ d_R\omega~=~(-1)^{|\omega|}d_L \omega, \tag{C}$$ where $\omega, ...


2

It must be something like $(-1)^{\text{parity of the permutation}}$, where the parity of the permutation can be defined (there are several possible definitions) as the parity of the number of transpositions into which the permutation is decomposed. One shows that the number of factors of any two such decompositions has the same parity. Another possible ...


2

The exterior algebra (generated by a vector space $V$) is an algebra. It is the quotient of the tensor algebra $TV$ by the relation $xy+yx=0$ for all $x,y\in V$. I think you are confusing what is the algebra and what is the vector space. $V$, in this context, does not form an algebra under the wedge product. However, the vector space $\Lambda(V)$ does. We ...


2

We do, but for historical reasons they are called (linear) isometries.


2

The set of differential forms along with addition, scalar multiplication, and the exterior product is an algebra (over a field). Add the exterior derivative and you get a differential algebra. That is, an algebra with an operation which is linear and obeys Leibniz's rule.


2

If $A$ is diagonalizable, so that there exists an invertible matrix $C$ such that $D=CAC^{-1}$ is diagonal, then $D=D^t=(CAC^{-1})^t=C^{-t}A^tC^t$. Since $D$ and $D^t$ have the same determinant, simply because the two matrices are in fact equal, it follows at once from this that $A$ and $A$ and $A^t$ have the same determinant. As diagonalizable matrices are ...


2

Fix a basis $e_i$ for $V$. Let $v = \Sigma c_{ij} e_i \otimes e_j$ be a tensor in your space. If $v$ is an -1 eigenvector, then $v + Tv = 0$. But $v + Tv = \Sigma (c_{ij} + c_{ji}) e_i \otimes e_j$, which implies that $c_{ij} = - c_{ji}$. Hence $v = \Sigma c_{ij} (e_i \otimes e_j - e_j \otimes e_i)$.


2

The cotangent bundle of $M \times F$, as a vector bundle, is the direct sum of the cotangent bundles of $M$ and $F$ (more precisely, of their pullbacks along the natural projections), so you're asking how to describe the exterior powers of a direct sum $V \oplus W$ of two vector bundles. The answer is that the exterior algebra is a graded tensor product ...


2

$\bigwedge^r T^*(M\times F)=\sum_{s+t=r}\bigwedge^s T^*M\otimes\bigwedge^t T^*F$


2

It is the second term that is always zero (because $d^2a = 0$), no matter what $k$ is. The first term is zero because $da$ has odd degree. In general $$ x \wedge y = (-1)^{|x||y|} y \wedge x$$ where $|x|,|y|$ are the degrees of $x,y$. In particular, if $x$ has odd degree then $x \wedge x = 0$.


2

You can definitely approach this from the perspective of exterior algebra. Indeed, the part of your question relating to your first identity has been asked before, so for the coordinate-free, exterior algebra proof of that, please take a look at my answer to the earlier question. So, let me know turn to your second identity. Before continuing, let me recycle ...



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