Let $V$ be an 3-dimensional vector space over $\mathbb{R}$. Let $\Lambda^3V^*$ denote the space of alternating trilinear forms on $V$. Note: An alternating trilinear form on $V$ is a map $\omega: V \times V \times V \to \mathbb{R}$ such that $\omega(v_1,v_2,v_3)=-\omega(v_2,v_1,v_3)=-\omega(v_3,v_2,v_1)$ and $\omega(av_1+v_4,v_2,v_3)=a\omega(v_1,v_2,v_3)+\omega(v_4,v_2,v_3)$.

(a) Show $\dim(\Lambda^3V^*)$=1.

(b) Given $L \in End(V)$ and $0 \neq \omega \in \Lambda^3V^*$ show that the function $\omega_L: V \times V \times V \to \mathbb{R}$ defined by $\omega_L(v_1,v_2,v_3):=\omega(Lv_1,Lv_2,Lv_3), \forall v_1,v_2,v_3 \in V$ is of the form $|L|\omega$ where $|L|$ is a constant which is dependent on $L$ but not on $\omega$.

(c) Show that in fact $|L|$ is the determinant of any matrix representing $L$ in terms of a basis of $V$. Is it true that $|LM|=|L||M|$ for any two determinants of endomorphisms $L,M$ of $V$?

(d) Is there way to define the trace of $L$ in a manifestly basis independent way similar to the definition of determinant in part (c)?

The following parts remain unsolved:

(e) We see the trace is a linear invariant of $L$, the determinant is a trilinear invariant of $L$. Now try to construct a bilinear invariant of $L$ along the same lines.

(f) If $L$ is diagonalisable, describe the last invariant in terms of eigenvalues of $L$.

(g) Given two linearly independent vectors $e_1$ and $e_2$ in $\mathbb{R}^3$, show that any non-zero $\omega \in \Lambda^3V^{*}$ may be used to explicitly construct a vector $e_3$ orthogonal to $e_1$ and $e_2$.

Note: For PART(e) I can't think of another invariant of $L$ , based on PART(f) this third invariant is related to eigenvalues of $L$ somehow, but what invariant besides the trace is closely related to the eigenvalues of $L$?

  • $\begingroup$ For (a), you are on a good way. Now use linearity, and think about what is $\omega(v,v,w)$. For (b), show that $\omega_L$ is also a skew-symmetric trilinear form, and then think about what it means that the dimension is 1. $\endgroup$ – celtschk Aug 4 '12 at 10:32
  • $\begingroup$ (a)-(d): yes. I recommend any introductory textbook on the multilinear algebra. $\endgroup$ – vesszabo Aug 4 '12 at 11:09
  • $\begingroup$ Hint: For (d): define $\omega_L(v_1,v_2,v_3)=Lv_1\wedge v_2\wedge v_3+v_1\wedge Lv_2\wedge v_3+v_1\wedge v_2\wedge Lv_3$. For (g): define $T_\omega:\mathbb{R}^2\to\mathbb{R}^\ast$ by $T_\omega(v_1,v_2)=\omega(v_1,v_2,\cdot)$, that is $T_\omega:\mathbb{R}^2\to\mathbb{R}^\ast$ by $T_\omega(v_1,v_2)v=\omega(v_1,v_2,v)$. $\endgroup$ – Yuki Aug 4 '12 at 13:15
  • $\begingroup$ @user31899: oopss... sorry!!! for (d): $\omega_L(v_1,v_2,v_3)=\omega(Lv_1,v_2,v_3)+\omega(v_1,Lv_2,v_3)+\omega(v_1,v_2,Lv_3)$... no idea why I wrote $v_1\wedge v_2\wedge v_3$... sorry!!! =p Try to write $\omega_l(v_1,v_2,v_3)=k\omega(v_1,v_2,v_3)$, for some $k$. $\endgroup$ – Yuki Aug 4 '12 at 22:56
  • $\begingroup$ @user31899: $T_\omega:(\mathbb{R}^3)^2\to(\mathbb{R}^3)^\ast$ (sorry, typo) is for PART(g), how can you canonically "move" from $(\mathbb{R}^3)^\ast$ to $\mathbb{R}^3$? For (e): you can do anything analogous you did in (b) ($\omega_L(v_1,v_2,v_3):=\omega(Lv_1,Lv_2,Lv_3)$), or in (c) ($\omega_L(v_1,v_2,v_3):=\omega(Lv_1,v_2,v_3)+\omega(v_1,Lv_2,v_3)+\omega(v_1,v_2,Lv_3)$) (try to "put $L$" in "two entries", and sum all the possibilities). For (f), (solve (e)) and note that "determinant" is the product of the eigenvalues and "trace" is its sum. $\endgroup$ – Yuki Aug 5 '12 at 1:32

About part e), you know both the trace and the determinant are coefficients of the characteristic polynomial of $L$, which in this case has degree $3= \dim V$. There is one coefficient you have not used yet, and perhaps this is the bilinear invariant you are looking for.


Thanks very much for the hints! I have now solved part (a)~(d). Following is my approach:


Let $L$ be the matrix of the operator L, based on the explicit formulas in PART(a) one has $ \omega_L(v_1,v_2,v_3)=\omega(Lv_1,Lv_2,Lv_3)$ $=$ $-\det(Lv_1,Lv_2,Lv_3).\omega(a,b,c)$ $=$ $-\det(L).\det(v_1,v_2,v_3).k_\omega$ $=$ $-\det(L).\omega(v_1,v_2,v_3)$. Therefore |L|=$\det(L)$ which is clearly independent of $\omega$.


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