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4

Yes. This is a corollary of Schur-Weyl duality. You need at least the additional assumption that your symmetric monoidal category is enriched over $k$-vector spaces. In general I don't see any reason to expect that the action of $k[S_n]$ is faithful; consider, for example, the special case where we only look at $1$-dimensional vector spaces. Sometimes. The ...


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

There does exist a notion of exterior powers of modules over arbitrary (commutative) rings, and they can be defined almost exactly the same way as for vector spaces. Here is a reference that goes through the details. Using this definition, the proof that determinants are multiplicative over a field generalizes straightforwardly to arbitrary rings. ...


3

Choose bases $(E_i)$ and $(F_j)$ for $A$ and $B$ and extend them to bases for $V$ and $W$, respectively. We get a basis $(E_i\otimes F_j)$ for $A\otimes B$, and in the same way the extensions give a basis for $V\otimes W$. Then, we get that $\bar{f}$ maps a basis of $A\otimes B$ to a linearly independent subset of $V\otimes W$ (a subset of a basis). This ...


2

Take a look at Couple stress theory for solids


1

To remove notational confusion, let $\lambda^1,\ldots,\lambda^n$ be the dual basis of $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$. Then, your goal is to show that $$d(x^i)=\lambda^i.$$ i.e. $d(x^i)=dx^i$, where the later is the formal symbol usually used for $\lambda^i$. Note that this justifies the notation $dx^i$. Now, compute: ...


1

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 ...


1

Given a $p$-form $\theta \in \bigwedge^p E^*$, we can define an alternating multilinear map $h \colon E^{n-p} \to \bigwedge^n E$ by $$h(u_1, \ldots, u_{n-p}) = \theta \wedge \tilde{u}_1 \wedge \ldots \wedge \tilde{u}_{n-p}.$$ Let $b \colon \mathbb{R} \to \bigwedge^n E$ be the linear map $$b(t) = t \omega.$$ Because $\bigwedge^n E$ is one-dimensional and ...


1

The equation you wish to prove is linear in each of the vectors $u_j$ and $v_j$. Therefore it suffices to show the identity when these vectors are basis vectors. There are $n$ basis vectors from which we now want to choose the $n$ vectors $u_1,\dots,u_n$. If we choose any two to be the same, then both sides of the identity vanish (and the identity is true), ...


1

Hint : (1) Let $$ U=[u_1\cdots u_n],\ V=[v_1\cdots v_n]$$ Then $$ (U^TV)_{ij} = u_i\cdot v_j =g(u_i,v_j) $$ (2) ${\rm det} (U^TV)={\rm det}\ U {\rm det}\ V$


1

As by the question linked by Michael, it is generally proved in a course about differential forms that $\alpha\wedge\beta = (-1)^{pq} \beta\wedge\alpha$ where $p$ and $q$ are the degrees of $\alpha$, $\beta$. Using this the result is immediate. But you could try and prove it, at least in your specific case. Hint: it should not be difficult to prove it for ...



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