Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a finite dimentional Euclidean space with the inner product $\langle...,...\rangle$, and let $k$ be an integer. Consider the polylinear form $X^k\times X^k\to{\mathbb R}$ $$ \big\langle x_1,...,x_k\; |\; y_1,...,y_k \big\rangle =\det\begin{pmatrix} \langle x_1,y_1\rangle & \dots & \langle x_1,y_k\rangle \\ \dots & \dots & \dots \\ \langle x_k,y_1\rangle & \dots & \langle x_k,y_k\rangle \end{pmatrix},\quad x_i,y_i\in X. $$ It is extended to a bilinear form $\langle...,...\rangle$ on the space $V_k(X)$ of polyvectors such that $$ \big\langle x_1\vee...\vee x_k\; |\; y_1\vee...\vee y_k \big\rangle =\det\begin{pmatrix} \langle x_1,y_1\rangle & \dots & \langle x_1,y_k\rangle \\ \dots & \dots & \dots \\ \langle x_k,y_1\rangle & \dots & \langle x_k,y_k\rangle \end{pmatrix},\quad x_i,y_i\in X. $$ Why is this bilinear form $\langle...,...\rangle$ on $V_k(X)$ positive definite?

I mean, from the properties of the Gram determinant it immediately follows that $\langle P|P\rangle>0$ for each non-zero elementary polyvector $P=x_1\vee...\vee x_k$. But why is the same true for all non-zero polyvectors, not necessary elementary?

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

I suppose what you call the space $V_k(X)$ of polyvectors means the $k$-th exterior power $\bigwedge^k X$ of $X$, and with $x_1\vee\cdots\vee x_k$ you mean $x_1\wedge\cdots\wedge x_k$.

Now $\bigwedge^k V$ is a vector space of dimension $\binom nk$ where $n=\dim X$, and if $e_1,\ldots,e_n$ is an orthonormal basis of $X$, then the elements $e_{i_1}\wedge\cdots\wedge e_{i_k}$, where $(i_1,\ldots,i_k)$ runs through the strictly increasing sequences in $\{1,\ldots,n\}^k$, for a basis of $\bigwedge^k V$. The bilinear form you describe is just the standard inner product with respect to this basis, and it is positive definite like any standard inner product is. Explicitly, if an element $\lambda\in V_k(X)$ is expressed in this basis, then (by bilinearity) $\left<\lambda,\lambda\right>$ is the sum of the squares of the $\binom nk$ coefficients, obviously non-negative.

share|improve this answer
    
Yes, what I ask is equivalent to what you say. But what do you mean by the standard inner product? –  Sergei Akbarov Jan 19 '13 at 11:02
    
@Sergei: By the standard inner product of a real vector space with respect to a given basis I mean the unique inner product for which that basis is orthonormal. For two vectors expressed in the basis, the inner product is the sum of the products of corresponding coordinates; you must have seen this defined soemwhere. –  Marc van Leeuwen Jan 19 '13 at 11:23
    
Ah, yes, I understood. If the basis $e_i$ is orthogonal, then the matrix of the bilinear form $\langle ...|...\rangle$ becomes diagonal. –  Sergei Akbarov Jan 19 '13 at 11:40
add comment

Take an orthonormal basis for $X$, say $e_1,\cdots,e_n$. Check that $$\{e_{i_1} \wedge \cdots \wedge e_{i_k} : 1 \leq i_1 < \cdots < i_k \leq n\}$$ gives an orthonormal basis for $X^k$ under that bilinear form. It's then clear that the bilinear form is positive definite when you write everything in terms of the basis and evaluate.

share|improve this answer
    
Sanchez, thank you also. If there were a possibility, I would accept your answer as well. –  Sergei Akbarov Jan 19 '13 at 12:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.