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
$X^i = \left(\begin{array}{x} x^1 \\\\ x^2 \\\\ x^3\end{array}\right)$
$X_i = \left(\begin{array}{ccc} x_1 & x_2 & x_3\end{array}\right)$
$X^i_j = \left(\begin{array}{ccc} x^1_1 & x^1_2 & x^1_3 \\\\
x^2_1 & x^2_2 & x^2_3 \\\\ x^3_1 & x^3_2 & x^3_3\end{array} \right)$
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.
Best Answer
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:
Finally, you may want to read: Basis vectors
By the way, I don't recommend to rely blindly on the picture given by matrices, specially when you are doing calculations.