I am reading a textbook on differential geometry and gauge theories and it it says off-hand that the alternating property of alternating multilinear forms implies that a form of this kind evaluated on a linearly dependent set of vectors is zero.

Would someone mind clarifying why the alternating property implies this? For example, let's say it's a 2-linear form which maps a pair of vectors in the same vector space to the reals, we switch the order of the pair in the Cartesian product and that changes the sign of the real which we map to.


If the two vectors are linearly dependent so there exists a weighted combination of them added together which adds to zero (or equivalent definition), why does that imply that evaluating the 2-form on that pair of vectors has to equal 0? I have studied differential forms and other exterior algebra, so any explanation in terms of these is fine, I would just like to know the proof for this.

EDIT: I did specify the multilinear case but to I will see if I can generalize to the trilinear case to answer the comment below, and then you just keep adding on indefinitely for a general case (can someone correct if this is wrong).


If any of these vectors equals 0 the form is 0 by multilinearity. From linear dependence, a vector in the set can be expressed in terms of the others, say

$v_3 = av_2 + bv_2$

$\phi(v_1,v_2,av_1 + bv_2)=-\phi(av_1 + bv_2,v_2,v_1)$

$\phi$ is trilinear so

$\phi(v_1,v_2,av_1 + bv_2) = a\phi(v_1,v_2,v_1) + b\phi(v_1,v_2,v_2)$

$-\phi(av_1 + bv_2, v_2, v_1)=-a\phi(v_1,v_2,v_1)-b\phi(v_1,v_2,v_2)$

$a\phi(v_1,v_2,v_1) + b\phi(v_1,v_2,v_2)=-a\phi(v_1,v_2,v_1)-b\phi(v_1,v_2,v_2)$

This implies that $\phi(v_1,v_2,v_3)=0$.

For the $k$th case:


From linear dependence:


$\phi(v_1,...,v_{k-1},av_1 +...+dv_{k-1})=-\phi(av_1+...+dv_{k-1},...,v_1)$

$\phi$ is $k$-linear so




Hence $\phi(v_1,...,v_{k-1},v_k)=0$.

  • $\begingroup$ You may also try to prove that they are actually equivalent. That is a covariant $k$-tensor $\alpha$ is alternating $\Leftrightarrow$ $\alpha(v_1,\dots,v_k) = 0$ whenever $k$-tuple $(v_1,\dots,v_k)$ is linearly dependent. $\endgroup$ – kelvinn aja Apr 5 '18 at 18:13
  • $\begingroup$ The answers so far handle the case of a $2$-form. Do you see how to generalize their answers to $k$-forms? $\endgroup$ – Jason DeVito Apr 5 '18 at 18:16
  • $\begingroup$ I've tried to generalize to the trilinear case, and for the k-linear case, you just add on indefinitely for $\phi(...,v_i,...,v_j,...)=-\phi(...,v_j,...,v_i,...)$ but should be the same argument as trilinear but with additional terms added. $\endgroup$ – Tom Apr 5 '18 at 18:56
  • $\begingroup$ I've tried for a $k$-form but not entirely sure if it's right, could you maybe point out where I have gone wrong? $\endgroup$ – Tom Apr 5 '18 at 19:43
  • $\begingroup$ The book is Gockeler and Schucker 'Differential Geometry, Gauge Theories and Gravity' if anyone is curious. $\endgroup$ – Tom Apr 14 '18 at 21:14

If either of the vectors are zero, then by bilinearity, ϕ($v_2$,$v_1$) = 0.

If $v_1$ and $v_2$ are linearly dependent nonzero vectors, then there must be some scalar c such that $v_1 = cv_2$. Then the statement


can be rewritten


But if ϕ is bilinear, then ϕ($v_2$,$cv_2$) = cϕ($v_2$,$v_2$) and ϕ($cv_2$,$v_2$) = cϕ($v_2$,$v_2$)

So we have

cϕ($v_2$,$v_2$) = -cϕ($v_2$,$v_2$)

Thus, ϕ($v_2$,$v_2$) = 0

  • $\begingroup$ The line "if $\phi$ is linear" contains typos ($1$ instead of $2$ or conversely), and more importantly: $\phi$ is bilinear, not linear. Linear would imply e.g. $\phi(av,av)=a\phi(v,v)$ instead of $a^2\phi(v,v)$. $\endgroup$ – Arnaud Mortier Apr 5 '18 at 18:05
  • $\begingroup$ Yes, I did specify multilinearity, but I get what he meant. $\endgroup$ – Tom Apr 5 '18 at 18:11

For any vector $v$, $$\phi(v,v)=-\phi(v,v)$$ Therefore (assuming the base field has characteristic different from $2$) $$\phi(v,v)=0.$$

Now if $v_2=\lambda v_1$, by bilinearity $$\phi(v_1,v_2)=\lambda\phi(v_1,v_1)=0.$$

  • $\begingroup$ Yes of course, I actually realized that linear dependence would imply that one could be expressed in terms of the other, hence you would effectively end up with the form evaluating the same vector apart from a factor, thanks for clarifying this. $\endgroup$ – Tom Apr 5 '18 at 17:57
  • $\begingroup$ @Tom You're welcome. Be careful though not to mix up linear and bilinear (as explain in comment to the selected answer). $\endgroup$ – Arnaud Mortier Apr 5 '18 at 18:06

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