Pure contravariant tensor

• Dec 16th 2009, 04:47 AM
Showcase_22
Pure contravariant tensor
A pure contravariant tensor of order (p,0) (the 0 shows that a vector space $\displaystyle V$ involved) has the form:

$\displaystyle \underbrace{ V^* \times \cdots \times V^* }_{p \ times} \rightarrow \mathbb{R}$

where $\displaystyle V^*$ is the dual space of $\displaystyle V$.

I'm having a lot of trouble trying to think of an example! I figure the easiest example is to take a vector space $\displaystyle \mathbb{R}^3$ and any $\displaystyle 3 \times 3$ matrix. These types of matrices will be in the dual space and they will represent a linear map.

To map a linear map to a scalar without using any vectors in $\displaystyle V$, I figure the easiest way is to just take the determinant of the matrix.

I think this example is okay, but I wondered if there was a better one I could use.

Anyway, my question is: Can anyone think of any other good examples of a pure contravariant tensor?
• Dec 16th 2009, 09:03 AM
NonCommAlg
Quote:

Originally Posted by Showcase_22
A pure contravariant tensor of order (p,0) (the 0 shows that a vector space $\displaystyle V$ involved) has the form:

$\displaystyle \underbrace{ V^* \times \cdots \times V^* }_{p \ times} \rightarrow \mathbb{R}$

where $\displaystyle V^*$ is the dual space of $\displaystyle V$.

I'm having a lot of trouble trying to think of an example! I figure the easiest example is to take a vector space $\displaystyle \mathbb{R}^3$ and any $\displaystyle 3 \times 3$ matrix. These types of matrices will be in the dual space and they will represent a linear map.

To map a linear map to a scalar without using any vectors in $\displaystyle V$, I figure the easiest way is to just take the determinant of the matrix.

I think this example is okay, but I wondered if there was a better one I could use.

Anyway, my question is: Can anyone think of any other good examples of a pure contravariant tensor?

fix an element $\displaystyle v \in V$ and define $\displaystyle \varphi: V^* \times \cdots \times V^* \rightarrow \mathbb{R}$ by $\displaystyle \varphi(f_1, \cdots , f_p)=\prod_{j=1}^p f_j(v).$ it's clear that $\displaystyle \varphi$ is mutilinear (and thus it induces a linear map $\displaystyle \tilde{\varphi}: V^* \otimes \cdots \otimes V^* \rightarrow \mathbb{R}$ defined by $\displaystyle \tilde{\varphi}(f_1 \otimes \cdots \otimes f_p)=\prod_{j=1}^p f_j(v)$).