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For an inner product space $(V,\langle \cdot,\cdot\rangle)$ the exterior algebra $\Lambda V$ inherits an inner product, which satisfies $\langle a_1\wedge \dots \wedge a_n,b_1\wedge\dots\wedge b_n\rangle = \det(a_i,b_j)$ on simple tensors of the same rank and is zero for tensors of different rank.

Question: Is the associated norm on $\Lambda V$ submultiplicative, i.e. is $\Vert a \wedge b\Vert\le\Vert a\Vert \cdot \Vert b\Vert$ for all $a,b\in \Lambda V$?


  • If $a$ and $b$ both have rank $1$, then this is true, because $\Vert a \wedge b \Vert^2 = \Vert a \Vert^2 \cdot \Vert b \Vert^2 - \langle a, b\rangle^2$.
  • If $a=a_1\wedge \dots \wedge a_n$ and $b=b_1\wedge\dots\wedge b_n$ are simple rank $n$ tensors ($\neq 0$), then $$\Vert a \wedge b\Vert^2 = \det\left (\begin{matrix} A & C \\ C & B \end{matrix}\right) = \det A \cdot \det B \cdot \det(I - B^{-1}C A^{-1}C),$$ where $A,B,C$ are $n\times n-$block matrices containing $\langle a_i,a_j\rangle$ and so on and the second inequaltiy is taken from this Wikipedia article on the determinant of block matrices. If the last determinant was $\le 1$, we'd be done, but I don't see how this would work.
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  • $\begingroup$ This appears to be equivalent to the question of whether the volume of an $n$-dimensional paralellopiped is less than or equal to the volume of a right angle parallelopiped with the same lengths, to which the answer is obviously "yes", which could be proved by induction on the number of factors, I guess. $\endgroup$ – ziggurism Jan 15 '18 at 14:37
  • $\begingroup$ @ziggurism Good intuition! This also goes through formally: Let $T$ be the orthogonal projection onto the subspace spanned by $b_1,\dots,b_n$ and replace $a_j$ by $\tilde a_j = a_j - Ta_j$. Then $\tilde C = 0$ and hence $\Vert a \wedge b \Vert = \Vert \tilde a \wedge b \Vert = \Vert (I-T) a \Vert \cdot \Vert b \Vert \le \Vert a \Vert \cdot \Vert b \Vert$. But what happens to nonsimple tensors? $\endgroup$ – Jan Bohr Jan 15 '18 at 15:31
  • $\begingroup$ for non-simple tensors, we have $\lVert a\wedge b+c\wedge d\rVert\leq \lVert a\wedge b\rVert+\lVert c\wedge d\rVert$ by triangle inequality. $\endgroup$ – ziggurism Jan 15 '18 at 15:52

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