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I'm trying to understand basic tensor analysis. I understand the basic concept that the valency of the tensor determines how it is transformed, but I am having trouble visualizing the difference between different valencies when it comes to higher order tensors.

I have this picture in my mind for the lower order tensors

$X^i = \left(\begin{array}{x} x^1 \\ x^2 \\ x^3\end{array}\right)$

$X_i = \left(\begin{array}{ccc} x_1 & x_2 & x_3\end{array}\right)$

$X^i_j = \left(\begin{array}{ccc} x^1_1 & x^1_2 & x^1_3 \\ x^1_2 & x^2_2 & x^3_2 \\ x^1_3 & x^2_3 & x^3_3\end{array} \right)$

for $X^{ij}$ and $X_{ij}$ they are represented in the same 2d array, but the action on a vector isn't defined in the same way as with matrices.

What I am having trouble with is intuitively understanding the difference between $X^{ijk}$, $X_{k}^{ij}$, $X_{jk}^{i}$, and $X_{ijk}$ (other permutations of the valence $(2,1)$ and $(1,2)$ omitted for brevity).

ADDED After reading the responses and their comments I came up with this new picture in my head for higher order tensors.

Since I am somewhat comfortable with tensor products in quantum mechanics, I can draw a parallel with the specific tensor space I'm used to.

If we consider a rank-5 tensor with a valence of (2,3) then we can consider it in the braket notation as

$ \langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle = X_{ijk}^{lm} $

Now if we operate with this tensor on rank-3 contravariant tensor, we are-left with a constant (from the inner product) and a rank-2 contravariant tensor, unmixed tensor product $\begin{eqnarray}(\langle \psi_i \mid \otimes \ \langle \psi_j \mid \otimes \ \langle \psi_k \mid \otimes \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle)(\mid \Psi_i \rangle \ \otimes \mid \Psi_j \rangle \ \otimes \mid \Psi_k \rangle) &=& c \mid \psi_l \rangle \ \otimes \mid \psi_m \rangle \\ &=& X_{ijk}^{lm}\Psi^{ijk} = cX'^{lm}\end{eqnarray}$

If we were to further operate with a rank-2 covariant tensor (from the right, per convention that a covector and vector facing each other is an implied direct product) we would simply get a number out.

One thing I am confused about though, is that in one of the answer to this question there was a point made that we are taking tensor products of a Vector space with itself (and possibly it's dual), however in the quantum mechanics picture (although I didn't rely on it in this example) we often take tensor products between different, often disjoint, subspaces of the enormous Hilbert space that describes the quantum mechanical universe. Does the tensor picture change in this case?

Any comments on my example would be appreciated.

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If $V$ and $W$ are different vector spaces, an element in the tensor product space $V^* \otimes W^*$ (for example) is a bilinear map which takes one vector from $V$ and one from $W$ and returns a scalar. You can still write mixed tensors like this in terms of components, if you introduce a basis $(e_1,\ldots,e_n)$ in $V$ and a basis $(f_1,\ldots,f_n)$ in $W$ and corresponding dual bases in $V^*$ and $W^*$. For example, $V \otimes W$ will have basis elements $e_i \otimes f_j$. Does that make things clearer? I don't know exactly what you mean by "tensor picture". –  Hans Lundmark Oct 28 '10 at 21:59
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2 Answers 2

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Since you asked for an intuitive way to understand covariance and contravariance, I think this will do.

First of all, remember that the reason of having covariant or contravariant tensors is because you want to represent the same thing in a different coordinate system. Such a new representation is achieved by a transformation using a set of partial derivatives. In tensor analysis, a good transformation is one that leaves invariant the quantity you are interested in.

For example, we consider the transformation from one coordinate system $x^1,...,x^{n}$ to another $x^{'1},...,x^{'n}$:

$x^{i}=f^{i}(x^{'1},x^{'2},...,x^{'n})$ where $f^{i}$ are certain functions.

Take a look at a couple of specific quantities. How do we transform coordinates? The answer is:

$dx^{i}=\displaystyle \frac{\partial x^{i}}{\partial x^{'k}}dx^{k}$

Every quantity which under a transformation of coordinates, transforms like the coordinate differentials is called a contravariant tensor.

How do we transform some scalar $\Phi$?

$\displaystyle \frac{\partial \Phi}{\partial x^{i}}=\frac{\partial \Phi}{\partial x^{'k}}\frac{\partial x^{'k}}{\partial x^{i}}$

Every quantity which under a coordinate transformation, transforms like the derivatives of a scalar is called a covariant tensor.

Accordingly, a reasonable generalization is having a quantity which transforms like the product of the components of two contravariant tensors, that is

$A^{ik}=\displaystyle \frac{\partial x^{i}}{\partial x^{'l}}\frac{\partial x^{k}}{\partial x^{'m}}A^{'lm}$

which is called a contravariant tensor of rank two. The same applies to covariant tensors of rank n or mixed tensor of rank n.

Having in mind the analogy to coordinate differentials and derivative of a scalar, take a look at this picture, which I think will help to make it clearer:

From Wikipedia:

alt text

The contravariant components of a vector are obtained by projecting onto the coordinate axes. The covariant components are obtained by projecting onto the normal lines to the coordinate hyperplanes.

Finally, you may want to read: Basis vectors

By the way, I don't recommend to rely blindly in the picture given by matrices, specially when you are doing calculations.

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Thanks for the response. I wasn't implying I use the matrix picture for calculations, it was just my intuitive way to understand what a mixed rank-2 tensor represents. –  crasic Oct 28 '10 at 19:24
    
Yes, I understood what you mean. It was just a side note. –  Robert Smith Oct 28 '10 at 20:13
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I prefer to think of them as maps instead of matrices. When you move to tensor bundles over manifolds, you won't have global coordinates, so it might be preferable to think this way.

So $x_i$ is a map which sends vectors to reals. Since it's a tensor, you're only concerned with how it acts on basis elements. It's nice to think of them in terms of dual bases: then $x_i(x^j)=\delta_{ij}$, which is defined as $1$ when $i=j$ and $0$ otherwise.

Similarly, $x^i$ is a map which sends covectors to reals, and is defined by $x^i(x_j)=\delta_{ij}$.

If you have more indices, then you're dealing with a tensor product $V^*\otimes\dotsb\otimes V^*\otimes V\otimes\dotsb\otimes V$, say with $n$ copies of the vector space and $m$ copies of the dual. An element of this vector space takes in $m$ vectors and gives you back $n$, again in a tensorial way. So, for example, $X_{ijk}$ is a trilinear map; $X^{ijk}$ is a trivector (an ordered triple of vectors up to linearity); $X_{ij}^k$ is a bilinear map taking two vectors to one vector; and so on.

It's worth thinking about these in terms of the tensors you've seen already. The dot product, for example, is your basic (0,2)-tensor. The cross product is a (1,2)-tensor. If you study Riemannian manifolds, it turns out you can use the metric to "raise and lower indices"; so the Riemannian curvature tensor, for example, is alternately defined as a (1,3)-tensor and a (0,4)-tensor, depending on the author's needs.

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I agree, except that you seem to have swapped upper and lower indices: $X^i$ usually denotes (the components of) a (1,0)-tensor, that is a vector, that is a map which sends covectors to numbers. And so on. (But in the last paragraph you are back to standard usage again.) –  Hans Lundmark Oct 28 '10 at 10:14
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Would it be fair to say that a (m,n) tensor $X_{*m}^{*n}$ takes n vectors to and m-tuple of vectors? What does this map do with covectors? –  crasic Oct 28 '10 at 10:38
    
Bleh, you're right. Indices bug me. I'll fix it. –  Paul VanKoughnett Oct 28 '10 at 10:38
    
An $(n,m)$ tensor takes $m$ vectors to an $n$-tuple of vectors. It would do the opposite on covectors, namely, take $n$ covectors to an $m$-tuple of covectors. –  Paul VanKoughnett Oct 28 '10 at 10:41
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@crasic and @Paul: It's not correct to say that an $(n,m)$ tensor takes $m$ vectors to an $n$-tuple of vectors -- if you think of it this way, the output is actually an $n$-vector (an element of the $n$-fold tensor product of $V$ with itself), which can't be identified with an $n$-tuple. When $n$ is bigger than 1, in many ways it's easier to think of an $(m,n)$ tensor as taking $m$ vectors and $n$ covectors and yielding a number that depends linearly on each input separately. –  Jack Lee Oct 28 '10 at 16:58
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