Abstract Nonsense

Crushing one theorem at a time

Alternating Forms of Maximal Degree


Point of post: In this post I discuss the topic of alternating forms of maximal degree as is discussed in Halmos’s section 31. I also give proof of the last theorem discussed in section 30.

Motivation

In this post we give two quick observations about alternating forms m-forms on an m-dimensional space. This will show, in essence that the space of alternating m-linear forms has dimension 1. So, the alternating forms make up a relatively small slice, loosely speaking, of the entirety of \text{Mult}_m\left(\mathscr{V}\right). The title is named as it is because we call a m-linear alternating form on an m-dimensional space an alternating form of maximal degree. The reason is that, as we shall see, if k>m the only k-linear alternating form is the zero form.

Theorem: Let m>0 and \mathscr{V} be a m-dimensional F-space. Then, if \text{Alt}_k\left(\mathscr{V}\right) is the set of all k-linear, skew-symmetric forms on \mathscr{V} then \text{Alt}_k\left(\mathscr{V}\right) is a subspace of \text{Mult}_k\left(\mathscr{V}\right).

Proof: Evidently \text{Alt}_k\left(\mathscr{V}\right) is non-empty since \bold{0}\in\text{Alt}_k\left(\mathscr{V}\right). So, now suppose that \alpha,\beta\in F and K,K'\in\text{Alt}_k\left(\mathscr{V}\right). Note then that if x\in\mathscr{V} and r,s\in[k] then

\displaystyle \begin{aligned}\left(\alpha K+\beta K'\right)(x_1,\cdots,\underbrace{x}_{r^{\text{th}}},\cdots,\underbrace{x}_{s^{\text{th}}},\cdots,x_k) &= \alpha K\left(x_1,\cdots,x,\cdots,x,\cdots,x_n\right)+\beta K'\left(x_1,\cdots,x,\cdots,x,\cdots,x_n\right)\\ &= \alpha 0+\beta 0\\ &= 0\end{aligned}

 

and since x,r and s were arbitrary it follows that \alpha K+\beta K' is alternating and thus in\text{Alt}_k\left(\mathscr{V}\right). \blacksquare

We now claim that \displaystyle \dim_F\text{Alt}_m\left(\mathscr{V}\right)\leqslant 1. To do this, we’ll show that any two alternating forms of maximal degree are linearly dependent.

Theorem: Let \mathscr{V} be a m-dimensional F-space with basis \{x_1,\cdots,x_m\} and K,K'\in\text{Alt}_m\left(\mathscr{V}\right). Then, K and K' are linearly dependent.

Proof: Note that since any field is a 1-dimensional vector space over itself it follows that the two scalars K\left(x_1,\cdots,x_m\right) and K'\left(x_1,\cdots,x_m\right) are linearly dependent and so there exists a non-zero scalar \alpha such that K(x_1,\cdots,x_m)=\alpha K'(x_1,\cdots,x_m). Note then by an earlier theorem we have that for any \displaystyle \sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\in\mathscr{V} it’s true that

\displaystyle \begin{aligned}K\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right) &= \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,j_k}K\left(x_1,\cdots,x_m\right)\\ &= \sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{m}\alpha_{k,\pi(k)}\alpha K'\left(x_1,\cdots,x_m\right)\\ &= \alpha\sum_{\pi\in S_m}\text{sgn}(\pi)\prod_{k=1}^{n}\alpha_{k,\pi(k)} K'\left(x_1,\cdots,x_m\right)\\ &= \alpha K'\left(\sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\right)\end{aligned}

and since \displaystyle \sum_{j_1=1}^{m}\alpha_{1,j_1}x_{j_1},\cdots,\sum_{j_m=1}^{m}\alpha_{m,j_m}x_{j_m}\in\mathscr{V} were arbitrary it follows that K=\alpha K' as desired. \blacksquare

We now prove that in fact, if \dim_F\mathscr{V}=m then \dim_F\text{Alt}_m\left(\mathscr{V}\right)=1 by showing the existence of a non-zero alternating m-form. We follow the proof given in Halmos closely:

Theorem: Let \mathscr{V} be a m-dimensional F-space and k\in[m]. Then, \text{Alt}_k\left(\mathscr{V}\right) is non-trivial.

Proof: We proceed by induction. For k=1 this is true since evidently \text{Hom}\left(\mathscr{V},F\right)=\text{Alt}_1\left(\mathscr{V}\right)\cong \mathscr{V}. Now, suppose that \text{Alt}_k\left(\mathscr{V}\right) for some 1\leqslant k<n  and thus has some non-zero K\in\text{Alt}_k\left(\mathscr{V}\right). Now since K\ne\bold{0} we may find x^0_1,\cdots,x^0_k\in\mathscr{V} such that K\left(x^0_1,\cdots,x^0_k\right)\ne 0 . So, now since k<n we may find some x^0_{k+1}\in\mathscr{V}-\text{span}\{x^0_1,\cdots,x^0_k\} and so evidently we may find some \varphi\in\text{Hom}\left(\mathscr{V},F\right) such that \varphi(x^0_1)=\cdots=\varphi(x^0_k)=0 and \varphi(x^0_{k+1})\ne 0 (for example, we may extend \{x^0_1,\cdots,x^0_{k+1}\} to some basis (since they must be l.i. since the value of an alternating multilinear form on a dependent set is zero) \mathscr{B} and so we may define \varphi to be the unique linear functional such that \varphi\left(\mathscr{B}-\{x^0_{k+1}\}\right)=\{0\} and \varphi(x^0_{k+1})=1).

We then define \tilde{K}\in\text{Mult}_{k+1}\left(\mathscr{V}\right) by

\displaystyle \tilde{K}(x_1,\cdots,x_{k+1})=\sum_{j=1}^{k}\left((j,k+1)K\right)(x_1,\cdots,x_k)\varphi(x_{k+1})-K(x_1,\cdots,x_k)\varphi(x_{k+1})\quad\mathbf{(1)}

We now claim that \tilde{K} is non-zero and alternating. The first fact is clear since

\tilde{K}\left(x^0_1,\cdots,x^0_k,x^0_{k+1}\right)=-K(x^0_1,\cdots,x^0_k)\varphi(x^0_{k+1})\ne 0

We suppose next that x_1,\cdots,x_k,x_{k+1} are vectors and r,s\in[k+1] (distinct) and x_r=x_s. Note that if x_r,x_s occur simultaneously in the slots of K in the expression for \mathbf{(1)} then that term is zero.

From here we break into two cases, namely s=k+1 and s\leqslant k. If the first is true then all the terms of \mathbf{(1)} vanish except

(r,k+1)K(x_1,\cdots,x_k)\varphi(x_{k+1})-K(x_1,\cdots,x_k)\varphi(x_{k+1})

and since x_r=x_{k+1} the transposition (r,k+1) does not change the value of K(x_1,\cdots,x_k) and thus the above vanishes. If j\leqslant k then the number of terms in \mathbf{(1)} which are non-vanishing is two, and since one of these two may be obtained by applying (r,s) to the other it follows that it vanishes as well. Thus, \tilde{K} is alternating and thus the induction is complete. \blacksquare

Corollary: If \mathcal{V} is a m-dimensional F-space, then \dim_F\text{Alt}_m\left(\mathscr{V}\right)=1

Proof: Since we proved any two m-forms must be linearly dependent it follows that \dim_F\text{Alt}_m\left(\mathscr{V}\right)\leqslant 1 and since, as was just proven, \text{Alt}_m\left(\mathscr{V}\right) is not trivial it follows that \dim_F\text{Alt}_m\left(\mathscr{V}\right)\geqslant 1. \blacksquare

References: Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

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November 15, 2010 - Posted by | Algebra, Halmos, Linear Algebra, Uncategorized | ,

1 Comment »

  1. […] the last entry. So, to be explicit what if we looked at some non-zero , which we know exists by prior theorem since . Then, we could find some (the superscripts are just meant so you don’t confused […]

    Pingback by Halmos Section 29,30 and 31 « Abstract Nonsense | November 18, 2010 | Reply


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