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1

It is convenient to use the Hodge star to simply the calculations. Choose a non-degenerate symmetric bilinear form $\left< \cdot, \cdot \right>$ on $V$ that has an orthonormal basis (for example, the one corresponding to the identity matrix) and let $(e_1, \ldots, e_n)$ be an orthonormal basis with respect to the chosen bilinear form. We will use ...

1

No, this is not true. For instance, take $n=k=2$, so $M=F[x]^2$, $\Lambda^k_{F[x]}M=F[x]$, and $\Pi:M^2\to F[x]$ is the map that sends $((a,b),(c,d))$ to $ad-bc$ (for $a,b,c,d\in F[x]$). Write $a_n$ for the degree $n$ coefficient of a polynomial $a\in F[x]$. Let $N=F$ and define $f:M^2\to N$ by $f((a,b),(c,d))=a_0d_1-b_1c_0$. Then $f$ is $F$-bilinear and ...

2

Let us write $Te_i = \sum_j a^j_i e_j$. For $1 \leq i < j \leq n$, we have $$(\Lambda^2(T))(e_i \wedge e_j) = Te_i \wedge Te_j = \left( \sum_{k_1} a_i^{k_1} e_{k_1} \right) \wedge \left( \sum_{k_2} a_j^{k_2} e_{k_2} \right) = (a_i^i a_j^j - a_i^j a_j^i) (e_i \wedge e_j) + \cdots$$ where the $\cdots$ don't involve $e_i \wedge e_j$ (as the coefficient ...

3

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 ...

2

Fix a scalar $\lambda\in F$. There is an evaluation map $m_\lambda:F[x]\rightarrow F$ such that $x\mapsto \lambda$. It induces a commutative diagram $\require{AMScd}$ \begin{CD} \Lambda^n_{F[x]} M @>{\Lambda^n_{F[x]}(1\otimes T-x\otimes \id_V)}>> \Lambda^n_{F[x]} M\\ @V{m_\lambda}VV @V{m_\lambda}VV \\ \Lambda^n_F V @>{\Lambda^n_F (T-\lambda ...

1

The problem is that your original formula for the Chern-Simons form is wrong. You should have the curvature $2$-form $F$ where you have $dA$. Those differ by a bracket term $A\wedge A$ (perhaps with some constant).

1

In $\mathbb{R}^3$, we can indeed more or less identify $u\times v$ with $u\wedge v$ for two vectors $u,v \in \mathbb{R}^3$ (more accurately, $u\wedge v$ is the linear form given by scalar product with $u\times v$; or another way of putting thigs it is that $u\wedge v$ gives an "axial vector" or "pseudovector", not a vector), but this no longer works for ...

2

The wedge product is defined to be the asymmetric tensors $$v_1 \wedge v_2 = v_1 \otimes v_2 - v_2 \otimes v_1$$ Therefore $e_1 \wedge e_1 = 0$ vanishes.

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 $$e_1\wedge e_1 = 0$$

2

We have $$e_1\wedge e_1=-e_1\wedge e_1,$$ so $e_1\wedge e_1=0$.

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 ...

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