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Let $L\colon \mathbb{R}^n \to \mathbb{R}^N$ be an injective linear map. By the Cauchy-Binet formula, $\det(L^TL)$ equals the sum of the squares of all minors of $L$ of order $n$: this looks just like a norm. Also, the square root of this number has a geometrical interpretation as the scale factor by which $L$ maps $n$-volumes, which makes Grassmann's exterior product come into mind.

Question If we let $\Lambda^n(\mathbb{R}^N)$ denote the vector space spanned by $n$-vectors

$$v_1\wedge \ldots \wedge v_n,\qquad v_j \in \mathbb{R}^N, $$

does there exist a scalar product $\langle ,\rangle$ on it such that $$\det(L^TL)=\langle Le_1\wedge \ldots \wedge Le_n, Le_1\wedge \ldots \wedge Le_n\rangle ?^{(\star)}$$

If the answer is affirmative, is this scalar product geometrically related to the concept of "oriented $n$-volume in $\mathbb{R}^N$"? And finally, is it possible to generalize all this to an arbitrary Riemannian manifold?

Bibliographical references as answers are fine. Thank you.


(*) $e_1\ldots e_n$ denotes the standard basis of $\mathbb{R}^n$.

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    $\begingroup$ See section 9.4 Differential Forms and Metrics in Jeff Lee's Manifolds and Differential Geometry. He explains precisely how an inner product on a space $V$ induces an inner product on the exterior algebra of $V$. I think you will find a lot of value in his presentation. $\endgroup$ Commented May 3, 2012 at 15:57
  • $\begingroup$ Thank you, I will look for this book. $\endgroup$ Commented May 3, 2012 at 16:08
  • $\begingroup$ @ItsNotObvious : Can you please make your comment into an answer? This way I can upvote and accept it. I find the reference you gave to be very useful. $\endgroup$ Commented May 31, 2012 at 20:59
  • $\begingroup$ Posted as answer. $\endgroup$ Commented May 31, 2012 at 21:08

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See section 9.4 Differential Forms and Metrics in Jeff Lee's Manifolds and Differential Geometry. He explains precisely how an inner product on a space V induces an inner product on the exterior algebra of V. I think you will find his presentation enlightening.

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