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This is homework so no answers please

The problem is: Find for which k, n, a k-alternating map $\omega$ can be written as $\omega=\omega_{1}\wedge...\wedge \omega_{k}$ were $\omega_{i}$ are alternating linear maps.

Attempt

It follows easily for k=1 and k=n. I also know it is decomposable for k=n-1. Now my issue is for $2\leq k\leq n-2$.

For $k=2$, the $e_{1}\wedge e_{2}+e_{3}\wedge e_{4}$ is not decomposable, which can be seen from assuming it is and then equating coefficients to get a contradiction for them.

For higher k, I suspect $\sum_{i=1}^{n} \wedge_{i\neq j}e_{i}$ written in increasing order (eg. $e_{1}\wedge e_{2}\wedge e_{3}+...+e_{n-2}\wedge e_{n-1}\wedge e_{n}$) should be non-decomposable.

However, showing that by solving the system of equations for the coefficients becomes more complicated; so I was wondering if there is a suggestion for a slick way to show they are decomposable (if they are indeed). Ideally a way not involving coefficients. For example is there a way to wedge things to $e_{1}\wedge e_{2}+e_{3}\wedge e_{4}$ and preserve non-decomposability?

Currently, to reduce the difficulty, I am trying induction on k for fixed n (base case is k=2).

Thanks

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3 Answers 3

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$\blacktriangleright$ For k=1 , it follows by definition.

$\blacktriangleright$ For $2\leq k\leq \lfloor \frac{n}{2}\rfloor$, we will show that $\omega:=e_{1}\wedge...\wedge e_{ \lfloor \frac{n}{2}\rfloor}+e_{\lfloor \frac{n}{2}\rfloor+1}\wedge...\wedge e_{ 2\lfloor \frac{n}{2}\rfloor}$ is non-decomposable (the floor notation was added to work for odd and even n).

We showed in class that for 1-covectors $\omega_{1}\wedge...\wedge \omega_{k}=0$ iff $\omega_{i}$ are linearly depends. Thus, if $\omega=\omega_{1}\wedge...\wedge \omega_{\lfloor \frac{n}{2}\rfloor}\Rightarrow \omega\wedge \omega=\omega_{1}\wedge...\wedge \omega_{\lfloor \frac{n}{2}\rfloor}\wedge \omega_{1}\wedge...\wedge \omega_{\lfloor \frac{n}{2}\rfloor}=0 $. However $\omega\wedge \omega= e_{1}\wedge...\wedge e_{2\lfloor \frac{n}{2}\rfloor}\neq 0$. Thus, $\omega $ is non-decomposable.

$\blacktriangleright$ didn't find a counterexample yet for For $\lfloor \frac{n}{2}\rfloor+1\leq k\leq n-2$.

$\blacktriangleright$ For k=n-1\ We will show that any $\omega\in \Lambda^{n-1}(V)$ is decomposable. To do that we will find basis covectors $\{v_{j} \}_{1}^{n}$ for $\Lambda^{n-1}(V)$ s.t. $\omega=\lambda v_{1}\wedge...\wedge v_{n-1}$. In other words, we want $\omega$ to be linearly dependent of other combinations of $\{v_{j}\}$.

The map $L:v\in\Lambda^{1}(V)=V \mapsto v\wedge \omega\in \Lambda^{n}(V)$ is linear by linearity of wedge product $(cv_{1}+v_{2})\wedge \omega=cv_{1}\wedge \omega+v_{2}\wedge\omega$. Thus, the rank-nullity theorem says $n=dim(V)=ker(L)+range(L)\Rightarrow ker(L)\geq dim(V)-dim(\Lambda^{n}(V))=n-1$ i.e. there exist (n-1) basis vectors $v_{1},..,v_{n-1}$ s.t. $v_{j}\wedge \omega=0$.

Since $dim(V)=n$ and $ker(L)\subset V$ is a subspace, there exists vector $v_{n}\in V$ s.t. $v_{1},...,v_{n}$ is a basis for V. Then, let $\eta_{j}:=v_{1}\wedge...\wedge \widehat{v_{j}}\wedge...\wedge v_{n}$ and so this is a basis for $\Lambda^{n-1}(V)$. Next we check that this is the desired basis. Since $\omega\in\Lambda^{n-1}(V)$,

$$\omega=\sum_{j}\lambda_{j}\eta_{j}\Rightarrow \text{for}~i<n,~ 0=\omega \wedge v_{i}=\lambda_{i}v_{1}\wedge...\wedge v_{n}\wedge v_{i}\Rightarrow \lambda_{i}=0.$$

Thus, $\omega=\sum_{j}\lambda_{j}\eta_{j}=\lambda_{n}\eta_{n}=\lambda_{n} v_{1}\wedge...\wedge v_{n-1}$ i.e. $\omega$ is decomposable.

$\blacktriangleright$ For $k=n$, by reindexing each basis vector to the increasing index one we get $\omega=\sum_{I}a_{I}e^{i_{1}}\wedge\cdots\wedge e^{i_{n}}=(ce^{1})\wedge\cdots\wedge e^{n}$ for some real constant c.

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  • $\begingroup$ For the second case, isn't there a 2 missing? i.e. $\omega\wedge \omega= 2 e_{1}\wedge...\wedge e_{2\lfloor \frac{n}{2}\rfloor}\neq 0$. $\endgroup$
    – JKEG
    Nov 12, 2019 at 23:24
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We assume that $V$ is a real vector space of dimension $n$ and that $1\leq k\leq n-1$. We also denote by $\pi :\wedge^k(V^*)\setminus\{0\}\to \mathbb{P}(\wedge^k(V^*))$ the projection onto the projectivization of $\wedge^k(V^*)$.

A nonzero $k$-covector $\omega\in \wedge^k(V^*)$ on $V$ is decomposable if and only if $\pi(\omega)$ is in the image of the Plücker embedding $\rho :Gr_k(V^\ast)\to\mathbb{P}(\wedge^k(V^*))$. Thus in order for every $k$-covector to be decomposable it is necessary and sufficient that $\rho$ be surjective. To check whether $\rho$ is surjective, we will look at the dimensions of the two manifolds.

  • If $\dim Gr_k(V^\ast)<\dim \mathbb{P}(\wedge^k(V^*))$, then $\rho$ is not surjective because a smooth map of constant rank is not surjective unless it is a smooth submersion.

  • If $\dim Gr_k(V^\ast)=\dim \mathbb{P}(\wedge^k(V^*))$, then $\rho$ is a local diffeomorphism, so it is open. On the other hand, $\rho$ is closed because Grassmannian is compact. Since $\mathbb{P}(\wedge^k(V^*))$ is connected, it follows that $\rho$ is surjective.

Therefore, $\rho$ is surjective if and only if $\dim Gr_k(V^\ast)=\dim \mathbb{P}(\wedge^k(V^*))$, that is, $k(n-k)= {{n}\choose{k}}-1.$ Since this equation is symmetric in $k$ and $n-k$, combining this with your observation on the case $2\leq k\leq \left \lfloor{\frac{n}{2}}\right \rfloor$, proves that this equation holds only for $k=1$ and $n-1$.

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Although this is quite an old question, I hope I might be able to provide some insight with what I consider to be a considerably simpler argument.

As you say, the $k=1$ and $k=n$ cases are both trivial, as well as the $k=0$ case if we care to include it. (Where a $0$-covector is simply a number.) If we limit ourselves to considering only basis $k$-covectors, the $k=n-1$ case follows when noticing that any $(n-1)$-covector must contain all but a single basis covector, and so a distinct pair must differ by only a single factor. Thus, using anticommutativity and bilinearity, we can shuffle the mismatched covectors to one end of their respective products and factor out the matching $(n-2)$-covector: \begin{align} &\varepsilon^{i_1} \wedge \dots \wedge \varepsilon^{i_k} \wedge \varepsilon \wedge \varepsilon^{i_{k+1}} \wedge \dots \wedge \varepsilon^{i_{n-2}} \\ &+ \varepsilon^{i_1} \wedge \dots \wedge \varepsilon^{i_l} \wedge \widetilde\varepsilon \wedge \varepsilon^{i_{l+1}} \wedge \dots \wedge \varepsilon^{i_{n-2}} \\ &= (-1)^{n-2-k}\varepsilon^{i_1} \wedge \dots \wedge \varepsilon^{i_{n-2}} \wedge \varepsilon \\ &+ (-1)^{n-2-l}\varepsilon^{i_1} \wedge \dots \wedge \varepsilon^{i_{n-2}} \wedge \widetilde\varepsilon \\ &= \varepsilon^1\wedge\dots\wedge\varepsilon^{i_{n-2}}\wedge((-1)^{n-2-k}\varepsilon + (-1)^{n-2-l}\widetilde\varepsilon). \end{align}

Since we're left with a linear combination of $1$-covectors, we've successfully decomposed the $(n-1)$-covector. (Of course, in general the matching factors could be permuted differently in the two products, but the only possible effect that could have would be to change the signs in the final linear combination, and the indices and exponents are getting hard enough to read as it is without cramming in more symbols. :P)

In this way, we see the $k=n-1$ case is in a sense dual to the $k=1$ case. This stems from the combinatoric nature of constructing basis $k$-covectors: choosing which collection of $(n-1)$-many covectors to wedge together in a product is the same as choosing which single covector to leave out of it.

By analogous reasoning, the $k=2$ case should be dual to the $k=n-2$ case. In spaces where $n > 3$, although some distinct pairs of $2$-covectors can share one common factor, it's not hard to see that there must be others that share none: $$ \varepsilon^{i_1} \wedge \varepsilon^{i_2} + \varepsilon^{i_3} \wedge \varepsilon^{i_4}. $$ And so we see $k=2$ does not always yield a decomposable $k$-covector. Similarly, although some distinct pairs of $(n-2)$-covectors can differ by only one factor, it's not hard to see that there must be others that differ by two. And so in such a case, our shuffling and factoring leaves us with an undecomposable linear combination of $2$-covectors, as above.

Continuing in this way, it becomes clear that the only values of $k$ for which $k$-covectors are always decomposable are $k=0,1,n-1,n$.

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