Suppose that $\phi_1,...,\phi_k\in V^*$ and $v_1,...,v_k\in V$, where $k=\dim V$. I am trying to show that $\phi_1\wedge\cdots\wedge \phi_k(v_1,...,v_k)=\frac{1}{k!}\det[\phi_i(v_j)]$, where $[\phi_i(v_j)]$ is a $k\times k$ matrix.

If the $\phi_i$ are linearly dependent, then the wedge product equals zero, and so we have the result. I am told to check the result for the dual basis in $V$. So suppose that $v_1,...,v_k$ form a basis for $V$ and that $\phi_1,...,\phi_k$ form the dual basis. The formula I have for the wedge product is $T\wedge S=\text{Alt}(T\otimes S)$, where $T$ is a $p$-tensor, $S$ a $q$-tensor, and $T\otimes S(v_1,...,v_p,v_{p+1},...,v_{p+q})=T(v_1,...,v_p)S(v_{p+1},...,s_{p+q})$, and where $\text{Alt}(T)=\frac{1}{p!}\sum_{\pi\in S_p}\text{sign}(\pi)T^\pi$, where $T^\pi=\text{sign}(\pi)T$.

Using these definitions, I have the following: Since each $\phi_i$ is a $1$-tensor, we do not have to worry about the sign, since every permutation will be the identity. So I have that $\phi_1\wedge\cdots\wedge\phi_k=\text{Alt}(\phi_1\otimes\cdots\otimes\phi_k)=\frac{1}{k!}(\phi_1\otimes\cdots\otimes\phi_k)$, and so $\frac{1}{k!}(\phi_1\otimes\cdots\otimes\phi_k)(v_1,...,v_k)=\frac{1}{k!}\phi_1(v_1)\cdots\phi_k(v_k)=\frac{1}{k!}$.

Is this computation correct? I am a little confused about what he matrix $[\phi_i(v_j)]$ is supposed to look like. The next part of the question asks me to "verify that the matrix does specify an alternating $k$-tensor on $V$, but I am not sure what this means. I know $\det$ is a tensor, but why is the matrix itself a tensor?


You computation is wrong, although you arrive to the correct result for the dual basis. Each $\phi_i$ is a 1-tensor, right, but you're computing $Alt$ of a tensor product of $k$ 1-tensors, which is a $k$-tensor, so you do have to take into account the sign of each permutation of $S_k$.

The matrix $[\phi_i(v_j)]$ is just the matrix

$ \begin{pmatrix} \phi_1(v_1) & \cdots & \phi_1(v_k)\\ \vdots & & \vdots\\ \phi_k(v_1) & \cdots & \phi_k(v_k) \end{pmatrix} $

If you take into account the sign of the permutations and recall the definition of the determinant in terms of permutations, you will get to the desired formula. In particular, for the dual basis, you get the identity matrix, whose determinant is $1$.

For your second question, I guess that the tensor specified by the matrix is the $\det$ tensor, so you have to verify that it is alternating. Another possibility is that, since every matrix defines a linear map, it defines a 1-tensor, but I consider this option unlikely since every linear map is trivially alternating.

  • $\begingroup$ Wouldn't that matrix just be the identity matrix? If so, shouldn't we still get the wedge product to be equal to $1/k!$? $\endgroup$ – confusedmath Apr 8 '18 at 19:36
  • $\begingroup$ And do you mean the sign of every permutation in $S_k$? $\endgroup$ – confusedmath Apr 8 '18 at 19:37
  • 1
    $\begingroup$ In general it's not the identity matrix, because $\phi_i$ is not specified to be dual to $v_i$. $\endgroup$ – Javi Apr 8 '18 at 19:38
  • 1
    $\begingroup$ I mean, in the formula for $Alt(\phi_1\otimes \cdots\otimes\phi_k)$. You've got a sum over $\pi\in S_k$, in which each term is multiplied by $sign(\pi)$, so yes, that's what I meant. $\endgroup$ – Javi Apr 8 '18 at 19:40
  • 1
    $\begingroup$ Oh, cool, I didn't read well, for the dual basis it would give you the identity matrix. $\endgroup$ – Javi Apr 8 '18 at 19:42

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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