Intuitive way to understand covariance and contravariance in Tensor Algebra

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

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

$X_i = left(begin{array}{ccc} x_1 & x_2 & x_3end{array}right)$

$X^i_j = left(begin{array}{ccc} x^1_1 & x^1_2 & x^1_3 \\
x^2_1 & x^2_2 & x^2_3 \\ x^3_1 & x^3_2 & x^3_3end{array} right)$

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.

edited Sep 5 '15 at 22:39

Sergio

289211

asked Oct 28 '10 at 9:46

crasic

2,13832228

2

$begingroup$
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".
$endgroup$
– Hans Lundmark
Oct 28 '10 at 21:59

$begingroup$
Isn't thinking of $X^i$ as an e.g. 3x1 vector and $X_i$ as a corresponding 1x3 vector a bit misleading? Although it makes sense for calculating $X_iX^i$ using the matrix form, the important relation between them is $X^i=g^{ij}X_j$, where $g$ is the metric tensor, right? So they're not generally tranposes of each other; it depends on what calculation you're doing, I think (e.g. this question on physics SE)
$endgroup$
– binaryfunt
May 22 '16 at 12:21

add a comment |