21joanna12 said:
Hi all,
I'm trying (and failing miserably) to understand tensors, and I have a quick question: is the inner product of a rank n tensor with another rank n tensor always a scalar? And also is the inner product of a rank n tensor with a rank n-1 tensor always a rank n-1 tensor that has been transformed/stretched in some way?
I would also be really grateful if anyone has a simple explanation of rank 2 tensors. I've looked up several posts on PF and elsewhere but I still don't really get them...
Thank you!
What do you mean by inner product? As the above posters have mentioned, there is no unique "inner product" for two general tensors. But I will expand on their answers a little bit.
Note first that WWGD specified that a rank 2 tensor is a bilinear map. To generalize his answer, a higher ranked tensor is simply a multi-linear map, linear in each argument.
Second, there are different kinds of tensors. There are the contravariant tensors, the covariant tensors, and the mixed tensors. The contravariant tensors are those which can be decomposed into sums of outer products of vectors (and therefore take one-forms as arguments and produces a scalar), the covariant tensors are those which can be decomposed into the sums of outer products of one-forms (and therefore take vectors as arguments and produces a scalar), and the mixed tensors are a mixture of the two.
In index notation, we denote the components of a contravariant tensor as all upper indices ##B^{ijk...}##, while we denote the components of a covariant tensor as all lower indices ##B_{ijk...}## and again a mixed tensor is a mixture of the two.
Generally speaking then, we need two different labels to declare a tensor. We need the contravariant rank as well as the covariant rank. So, we usually say a tensor is of rank (n,m). For example, a (2,0) tensor would be a rank 2 fully contravariant tensor. A (0,3) tensor would be a rank 3 fully covariant tensor, and a (2,4) tensor would be a rank 6 mixed tensor.
Because you have specified that you are looking for an "inner product", then that necessarily means you have a metric tensor present on the manifold. The metric tensor explicitly can give the inner product between two vectors ##g(V,W)\equiv \left<V,W\right>=g_{ij} V^i W^j##. Because the metric tensor can do this, and since the metric tensor defines an inner product, it also defines an isomorphism between the tangent space (space of vectors) and the cotangent space (space of one forms) ##\tilde{B}\equiv g(B,\quad)##, or in index notation ##B_i = g_{ij}B^j##. Because of this isomorphism defined, we can use the metric to raise and lower indices of tensors ##B_i^{jkl...}=g_{im}B^{mjkl...}##, and therefore, define contractions between two different tensors.
Say you have tow tensors ##A## and ##B## of type (n,0) and (m,0) respectively where ##m\leq n##. Then you can contract m of the indices between the two tensors by using the metric ##g_{ij}g_{kl}...A^{ik...}B^{jl...}##, where there is one factor of g for each index that is being contracted (so there are m factors, for the m indices on the tensor B). You can perhaps think of this as some type of extension of an inner product. It sounds like, from your post, that this is what you're thinking of. But beware that this is not usually called "THE inner product" between two tensors.
In this scenario, if you have two rank n tensors, then indeed the result will be a scalar because in that case you have ##g_{ij}g_{kl}...A^{ik...}B^{jl...}=A_{jk...}B^{jk...}##, if you have a rank n tensor and a rank n-1 tensor and this process is what you did, then you will end up with a vector, NOT an n-1 tensor. There will be 1 left over index that you can't contract.