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I'm trying to understand tensor analysis from a more modern point of view using Spivak's Calculus on Manifolds. I find myself getting very confused when I try to relate the modern viewpoint to my previous exposure to old-fashioned expositions of Cartesian tensors. For example, the conductivity tensor relates the current vector field to the electric vector field in an anisotropic medium: $$\bf{J}=\sigma \bf{E}$$ From my previous exposure I'd call $\sigma$ a rank 2 tensor that maps the electric field at a point $p$ to the current field at the same point, i.e.: $$\sigma: \bf{E} (p) \rightarrow \bf{J} (p)$$ Which I guess is really just $$\sigma: \mathbb{R}^3\rightarrow\mathbb{R}^3$$ But in Spivak, a tensor $T$ of rank 2 is a multi-linear mapping from two copies of some vector space $V$ to the reals: $$T: V \times V \rightarrow \mathbb{R}$$

Are these equivalent descriptions of the same mathematical object? If so, can you help me see the equivalency?

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  • $\begingroup$ Yes. They are equivalent descriptions. $\endgroup$
    – mike stone
    Commented Jan 29, 2022 at 23:15
  • $\begingroup$ Thanks @mikestone , that's was my guess. However, a simple confirmation that my guess was right isn't actually very helpful. Any chance you could provide an answer outlining the connection between the two descriptions? In the case of the conductivity tensor, what is the vector space $V$? $\mathbb{R}^3$? $\endgroup$ Commented Jan 29, 2022 at 23:19
  • $\begingroup$ Further browsing makes me think that the accepted answer to this question on the Math StackExchange is what I need to understand. Particularly the comment that every matrix $M$ defines a binlinear map $x^T M y$ mapping $V \times V$ to $\mathbb{R}$. Am I on the right track? $\endgroup$ Commented Jan 29, 2022 at 23:40
  • $\begingroup$ I'll add an answer. $\endgroup$
    – mike stone
    Commented Jan 30, 2022 at 13:02

3 Answers 3

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There is a difference between a matrix that defines a linear map $$ M:V\to V, \quad {\bf y}={\bf M}{\bf x}, \quad y^i={M^i}_j x^j $$ and one that defines a quadratic form $$ {\bf x}^T{\bf M} {\bf x}= x^i M_{ij}x^j $$ I suggest you look at section 10.2.2 (p 393) of our book for more details.

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  • $\begingroup$ Thank you. Section 10.2.2 look like just what I needed. $\endgroup$ Commented Jan 30, 2022 at 18:40
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The key thing to keep in mind is that vectors themselves are (linear) maps: $$v: \mathbb{R}^n \to \mathbb{R} \,.$$ If you plug a single vector $v \in \mathbb{R}^n$ into a tensor $T: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$, you are left with a map $$Tv: \mathbb{R}^n \to \mathbb{R} \,,$$ which is by definition also a vector. In that way, you can interpret a rank 2 tensor as either a map from two vectors to $\mathbb{R}$ or as a map from one vector to another. Remember that the space of linear maps on $\mathbb{R}^n$ is again $\mathbb{R}^n$, which is why the vector $v$ can be though of as an element of either.

Note that I cheated a bit and identified all vector spaces with $\mathbb{R}^n$. This is only possible once you pick bases in your spaces. In reality you also need differentiate between a vector space and its dual space. But as long as you work in a metric space, this distinction is not relevant, you can always use the metric tensor to move between the spaces.

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There are three equivalebt notions of tensors.

  1. Transformational - this is usually discussed in textbooks dealing with general relativity. They are written with imdices such as $T_{ijk}$

  2. Algebraic - this is usually dealt with in mathematics books in abstract algebra. They are written $u \otimes v \otimes w$

  3. Multilinear - these are usually discussed in more modern books dealing with general relativity.

Unfortunately, there are few books that discuss all three in a coherent and pedagogic fashion.

Your example of $T: V \times V \rightarrow \mathbb{R}$ can be transformed into $T: V \rightarrow V^*$. The rule is you can transfer a vector space factor from one side to the other by decorating with a dual. If there are no factors left, then you leave the ground field, which in this case is $\mathbb{R}$.

Thus your $\sigma:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ is equivalently, $\sigma:\mathbb{R}^3 \times (\mathbb{R}^3)^* \rightarrow \mathbb{R}$. This shows that it is a mixed tensor of type (1,1).

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