Inner product of two tensors

[Rick16]

TL;DR

Why can't the inner product of two tensors be a scalar?

I am trying to learn tensor calculus with Bernacchi's book "Tensors made easy". For the inner product of two rank-2 tensors he gets four different results, each of them a rank-2 tensor. Why can't there be a fifth solution, a scalar? What is wrong with the following equation?
$$\mathbf A \cdot \mathbf B = (A_{\mu\nu} \tilde e^\mu \tilde e^\nu)\cdot(B^{\alpha\beta}\vec e_\alpha \vec e_\beta)=A_{\mu\nu}B^{\alpha\beta} \tilde e^\mu \tilde e^\nu\vec e_\alpha \vec e_\beta=A_{\mu\nu}B^{\alpha\beta}\delta^\mu_\alpha\delta^\nu_\beta=A_{\alpha\beta}B^{\alpha\beta}=C,~i.e.~a~scalar$$

 

[Orodruin]

I do not have the book you are referring to, but I would assume they introduce an inner product as a single contraction between two tensor indices. In that case, your scalar is not a single, but a double, contraction. That does not mean it may not be an interesting quantity though.

Do note that the full contraction is also not unique unless one of the tensors is symmetric. You have a choice in which indices you contract with each other.

 

[Rick16]

So there is nothing wrong with the double contraction? Bernacchi writes about the inner tensor product "Rank $\geq$ 2 tensors leave residual indexes". This looked to me like they always must leave residual indexes. I guess he had only single contractions in mind and I misunderstood that he meant inner products in general.

 

[Orodruin]

No, there is nothing wrong with double contractions. They can often be useful. Consider for example the Lagrangian of the electromagnetic field, which contains the term $F_{\mu\nu}F^{\mu\nu}/4$ where $F$ is the electromagnetic field tensor.

 

[Rick16]

Thank you very much for clarifying this.

 

[Chestermiller]

TL;DR Summary: Why can't the inner product of two tensors be a scalar?

I am trying to learn tensor calculus with Bernacchi's book "Tensors made easy". For the inner product of two rank-2 tensors he gets four different results, each of them a rank-2 tensor. Why can't there be a fifth solution, a scalar? What is wrong with the following equation?
$$\mathbf A \cdot \mathbf B = (A_{\mu\nu} \tilde e^\mu \tilde e^\nu)\cdot(B^{\alpha\beta}\vec e_\alpha \vec e_\beta)=A_{\mu\nu}B^{\alpha\beta} \tilde e^\mu \tilde e^\nu\vec e_\alpha \vec e_\beta=A_{\mu\nu}B^{\alpha\beta}\delta^\mu_\alpha\delta^\nu_\beta=A_{\alpha\beta}B^{\alpha\beta}=C,~i.e.~a~scalar$$

Shouldn't that be the double-dot product?

 

[Rick16]

Shouldn't that be the double-dot product?

Yes, I wrote that wrong. The way I wrote the 3rd term, it looks like an outer product. I guess it should be written $A_{\mu\nu}B^{\alpha\beta}\tilde e^\mu \tilde e^\nu(\vec e_\alpha \vec e_\beta)$.