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9

I think the article you quote is using slightly facetious language in order to seem more interesting. In particular, when it seems to frame its question as "what are some other definitions of $i$?", it would be more honest to say What are the ways to define a multiplication operation on $\mathbb R^2$ which is compatible with normal vector addition, in a ...


8

This is in general not true. One easy way to see this, is the following: Assume $V$, $W$ are $n$ and $m$ dimensional vector fields over the complex numbers, then it is fairly easy to show that $V\otimes W$ is isomorphic to $\mathbb{C}^{n\times m}$ with the following isomorphism $\phi: V\otimes W \rightarrow \mathbb{C}^{m\times n}$, which is defined as ...


7

1) The equation $i^2=iq+p$ is not a recursive definition. It is defined as a solution of a quadratic equation. It is just like our normal definition $i^2=-1$. In this definition you can solve $i$ to get the two solutions, just like when you "solve" $i^2=-1$, you get our ordinary definition of $i$. 2) This is to say, when you solve $i^2=iq+p$, i.e., ...


6

Lets assume that $\dim V=m$ and $\dim W=n$ with $m\geq 2$ and $n\geq2$. Suppose that $\{v_1,v_2\}$ are linearly independent in $V$ and that $\{w_1,w_2\}$ are linearly independent in $W$. Seeking a contradiction, suppose that $$ v_1\otimes w_1+v_2\otimes w_2=v\otimes w\tag{1} $$ Then extend $\{v_1,v_2\}$ to a basis $\{v_1,v_2,v_3,\dotsc,v_m\}$ of $V$ and ...


4

It is necessarily possible to write all tensors as a simple tensor when either $V$ or $W$ has dimension $1$ (or $0$). I will leave this to you to confirm, the proof is fairly trivial. On the other hand, suppose that $V$ and $W$ have dimensions $2$ or greater, and that we are given a basis of vectors $v_i$ and $w_i$ for the respective spaces. Then I am ...


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 think the generalized complex number systems can simply be defined as rings of the form $\mathbb R[X]/(f(X))$ with $f(X)$ a real quadratic polynomial and $i$ is a notation of $\overline X$. For the meaning of this symbols maybe you can refer to wiki.


4

As Darij remarked in his comment, if $R$ is not a ${\mathbb Q}$-algebra one cannot argue that $\bigwedge^n P$ and ${\mathfrak S}^n P$ are summands of $P^{\otimes n}$, thereby reducing the statement to showing that projectivity and flatness are stable under taking tensor powers. Instead, one might argue as follows: Freeness: If $F$ is a free $R$-module on ...


4

I think there is an importan point that has been overlooked in the above answers: The exterior derivative is the only linear natural operator in the list. This is explained with several variations the book by Kolar, Michor and Slovak cited in Yuri Viatkin's answer. The Lie derivative is also natural under general diffeomorphisms but only as a bilinear ...


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

The answer to this question turned out to be rather uninteresting. There was an error on one of the Wikipedia articles which I had consulted, and this lead to my confusion. By the way, having done a bit more reading, I think that using the Pl├╝cker relations the way that I did, i.e. to decide when an element $\alpha \in \Lambda^2(X)$, $X$ a ...


3

A $k$-vector $w \in \bigwedge^kV$ is $m$-decomposable if there is a linearly independent set $\{e_1, \dots, e_m\}$ of $V$ and $\alpha \in \bigwedge^{k-m}V$ such that $w = e_1\wedge\dots\wedge e_m\wedge\alpha$; note that $k$-decomposable is what is normally called decomposable. Furthermore, $w$ is strictly $m$-decomposable if it is $m$-decomposable but not ...


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

Like GFR, I also have no interest in an extended discussion. This post is meant to be an extended comment for the sake of adding perspective. Personally, I like to define the exterior derivative $d$ in a coordinate-free, invariant way: Fact/Def: Let $M$ be a smooth manifold (possibly with boundary). Then there are unique operators $d \colon ...


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

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 ...


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

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

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.


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

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


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

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

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

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

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

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

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 $v = \Sigma c_{ij} (e_i \otimes e_j - e_j \otimes e_i)$.



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