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For the last few days, I have been trying to answer the following algebraic question in exterior algebra. The following question appears as an algebraic step in the context of existence of solutions of a certain system of PDE. I have asked a special case of the problem in

Link: Inequalities Involving Wedge Product (Reference Request)

with the hope that this particular case will do what I have in mind, but this did not turn out to be the case. Any help in this direction is welcome.


Let $N$ be a linear subspace of $\Lambda^2(\mathbb{R}^n)$ satisfying $$ \omega\wedge\omega\neq 0,\text{ for all }\omega\in N,\omega\neq 0. $$ Is it true that there exists a $a\in\Lambda^4(\mathbb{R}^n)$, $a\neq 0$ such that $$ \langle a;\omega\wedge\omega\rangle>0,\text{ for all }\omega\in N,\omega\neq 0? $$ Comment: My guess is that there will be such an $a$. But I could not prove it.

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This seems to be false for $n=4$: take $\omega_{\pm}=dx_1\wedge dy_1\pm dx_2\wedge dy_2$; then $\omega_+\wedge\omega_+=-\omega_-\wedge\omega_-$. – algori Nov 20 '12 at 17:56
Note that, your example is not the right one. For $\omega_{+}$,$\omega_{-}$ both cannot be in $N$ as $\omega_{+}+\omega_{-}=2dx_1\wedge dy_1$ which is decomposable whereas $N$ cannot contain non-zero decomposable forms. – tatin Nov 20 '12 at 18:29
tatin -- when you say "Let $N$ be a subspace" do you mean a vector subspace? – algori Nov 20 '12 at 18:47
Yes. A vector subspace. – tatin Nov 20 '12 at 18:50
I have modified so to avoid any confusion. – tatin Nov 20 '12 at 18:52
up vote 16 down vote accepted

No, such an $a$ does not always exist. The lowest dimension in which an example without an $a$ could possibly exist is dimension $5$, and, low and behold, there is such an example. I will give one now, but I will also leave the $8$-dimensional example as part of my answer, since I think it is nice and since it illustrates a general principle.

Example 1: For the $5$-dimensional example, let's identify $\mathbb{R}^5$ with the space $\mathsf{S}$ of traceless, symmetric, $3$-by-$3$ matrices with real entries. For each skew-symmetric $3$-by-$3$ matrix $x$, define a $2$-form $\omega_x$ on $\mathsf{S}$ by the rule $$ \omega_x(s_1,s_2) = \mathrm{tr}\bigl(\ x\ [s_1,s_2]\ \bigr) = \mathrm{tr}\bigl(\ x\ (s_1s_2-s_2s_1)\ \bigr) $$ for all $s_1,s_2\in\mathsf{S}$. The space of such $2$-forms is a $3$-dimensional vector space $N\subset\Lambda^2(\mathsf{S}^\ast)$. It is easy to check that ${\omega_x}^2\not=0$ for all nonzero $x$. However, the cone in $\Lambda^4(\mathsf{S}^\ast)$ consisting of the squares of elements of $N$ cuts every hyperplane in $\Lambda^4(\mathsf{S}^\ast)$, so it does not lie properly on one side of any hyperplane. Thus, there does not exist an $a\in\Lambda^4(\mathsf{S}^\ast)$ such that $\langle a; {\omega_x}^2\rangle>0$ for all nonzero $x$.

A word about the proof: This can be done by hand, just writing everything out in a basis, but there is an easier way: Note that the entire construction is invariant under action by $\mathrm{O}(3)$, i.e., if $A$ is an orthogonal $3$-by-$3$ matrix, then, letting $A$ act on $\mathsf{S}$ by $A\cdot s = A s A^{-1}= AsA^{\mathsf{T}}$ and letting $A$ act on $N$ by $A\cdot \omega_x = \omega_y$ where $y = AxA^{\mathsf{T}}$, one sees that the action of $A$ on $S$ preserves the subspace $N$, acting transitively on the unit sphere in $N$, so either ${\omega_x}^2$ vanishes identically or it is never zero for nonzero $x$. Computing one example shows that the latter holds. Next, the cone of squares of elements of $N$ in $\Lambda^4(\mathsf{S}^\ast)$ must also be invariant under the action of $\mathrm{O}(3)$, and, if it there were an $a$ in this space whose inner product with all of the elements of the cone were nonnegative, then averaging $a$ over the action of $\mathrm{O}(3)$ would yield a nonzero vector $\bar a$ in $\Lambda^4(\mathsf{S}^\ast)$ that was fixed under $\mathrm{O}(3)$. However, $\Lambda^4(\mathsf{S}^\ast)$ is isomorphic to $\mathsf{S}$ as an $\mathrm{O}(3)$-module and hence is irreducible. In particular, $\mathrm{O}(3)$ does not fix any nonzero vector in $\Lambda^4(\mathsf{S}^\ast)$.

Added remark: This example generalizes to all of the irreducible representations $H_k$ of $\mathrm{SO(3)}$, where the dimension of $H_k$ is $2k{+}1$.

Example 2: Here is an example of an $8$-dimensional subspace $N\subset\Lambda^2(\mathbb{R}^8)$ that has no corresponding $a\in\Lambda^4(\mathbb{R}^8)$.

Recall that the compact Lie group $\mathrm{SU}(3)$ has dimension $8$ and its de Rham cohomology is nontrivial only in degrees $0$, $3$, $5$, and $8$. Identify $\mathbb{R}^8$ with ${\frak{g}} = {\frak{su}}(3)$, the tangent space of $\mathrm{SU}(3)$ at the identity matrix, and note that, if $\phi\in\Omega^3\bigl(\mathrm{SU}(3)\bigr)$ is a bi-invariant $3$-form representing a nonzero element of $H^3_{dR}\bigl(\mathrm{SU}(3)\bigr)$, then the value of $\phi$ at the identity is the Cartan $3$-form $$ \kappa(x,y,z) = -\beta\bigl(x,[y,z]\bigr), $$ where $\beta:{\frak{g}}\times{\frak{g}}\to\mathbb{R}$ is the Killing form (which is nondegenerate and allows us to identify $\frak{g}$ with $\frak{g}^\ast$ in a natural way). Now set $$ N = \{i_x\kappa\ |\ x\in \frak{g} \}\subset \Lambda^2(\frak{g}^\ast)\simeq\Lambda^2(\mathbb{R}^8), $$ where $i_x$ denotes interior product with $x\in\frak{g}$. Then $N$ has dimension $8$, and it is easy to see, from the multiplication properties of the Lie algebra ${\frak{g}} = {\frak{su}}(3)$, that $\omega^2\not=0$ for any nonzero $\omega\in N$.

Now, if there were an $a\in \Lambda^4(\mathbb{R}^8)\simeq\Lambda^4({\frak{g}}^\ast)$ such that $$ \langle a; \omega^2\rangle >0 $$ for all nonzero $\omega\in N$, then, letting $\Lambda^4(ad^\ast):\mathrm{SU}(3)\to \mathrm{Aut}\bigl(\Lambda^4({\frak{g}}^\ast)\bigr)$ be the induced representation on $4$-forms by the coadjoint representation of $\mathrm{SU}(3)$ on $\Lambda^4({\frak{g}}^\ast)$, one can take an average $$ \bar a = \int_{\mathrm{SU}(3)} \Lambda^4(ad^\ast)(g)(a)\ dg, $$ (where $dg$ denotes the bi-invariant Haar measure on $\mathrm{SU}(3)$). Because the pairing $\langle,\rangle$ is invariant under $\Lambda^4(ad^\ast)$, it follows that $\bar a$ must be invariant under $\Lambda^4(ad^\ast)$ and it must satisfy $$ \langle \bar a; \omega^2\rangle >0 $$ for all nonzero $\omega\in N$, implying that $\bar a$ is a nonzero element of $\Lambda^4({\frak{g}}^\ast)$ that is fixed under the action of $\mathrm{SU}(3)$. However, because $H^4_{dR}\bigl(\mathrm{SU}(3)\bigr)= (0)$ (as already noted), this is impossible.

Obviously, this same technique is going to work for any compact semisimple Lie group of dimension $8$ or more, so there are lots of examples.

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Thanks again. Thank a lot. I'm not familiar with representation theory. It will take me some time to understand your counterexample fully. Is there any direct counterexample that constructs the subspace explicitly? – tatin Nov 21 '12 at 4:09
@tatin: Well, I woke up this morning and realized that there actually is an example in dimension $5$. (Obviously, dimension $4$ is impossible.) I'll amend my answer above to include the example in dimension $5$. It's simple enough that you should be able to write it out and check it by hand without representation theory (which the $8$-dimensional example is not). – Robert Bryant Nov 21 '12 at 12:43
Thanks a lot! I do not know how to thank you enough!!! – tatin Nov 21 '12 at 18:02

For what is worth, here is a simple coordinate example in dimension 6. Take $$ \omega_0 = dx_1\wedge dx_2 + dx_3\wedge dx_4 + dx_5\wedge dx_6 , $$ $$ \omega_1 = dx_1\wedge dx_3 + dx_2\wedge dx_4 , $$ $$ \omega_2 = dx_3\wedge dx_5 + dx_4\wedge dx_6 , $$ $$ \omega_3 = dx_5\wedge dx_1 + dx_6\wedge dx_2 . $$ The sum of squares of these forms is zero, but the square of any nontrivial linear combination is not.

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@Sergei: That's a nice example. It, too, has a lot of symmetry since it is invariant under the diagonal action of $\mathrm{O}(3)$ when you split it as $(x_1,x_3,x_5)\oplus(x_2,x_4,x_6)$. That plus the obvious scaling and exchange makes a group of dimension $4$ that acts irreducibly on $\mathbb{R}^6$. Is there a larger subgroup of $\mathrm{GL}(6)$ that preserves this subspace of $\Lambda^2(\mathbb{R}^6)$? – Robert Bryant Nov 22 '12 at 2:25
@Sergei: Thank you. Very nice example!!. – tatin Nov 22 '12 at 6:59

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