Can the following tensors be contracted?

This is an assignment my professor gave me this summer: Can the following tensors be contracted?

1. $A_i A^i B^{ jrm} B_{jrm}$

2. $A_j A^i B^{ jrm} B_{irm}$

3. $A_i A^j B^{ irm} B_{jrm}$

4. $A_r A^i B^{ rmj} B_{imj}$

• Could you elaborate on the format of your question? Is this supposed to be multiple choice? Where is this question coming from... some context would make this a better question. – James S. Cook Jul 18 '13 at 21:25
• I'm assuming the Einstein summation convention us used here. Correct? – Robert Lewis Jul 18 '13 at 21:27
• Thanks for the comment... OK, I would like to know if I have successfully combined indices on the pairs A and B in a way that they could be contracted. And if they can, do they combine into one tensor, like the ones I list below? The contracted tensor for 1 is a, 2 is b, 3 is c, and 4 is d – CuriousGeorge119 Jul 18 '13 at 21:29
• My professor mentioned Einstein summation, yes, although I am not very familiar with it – CuriousGeorge119 Jul 18 '13 at 21:30
• @CuriousGeorge119 I replaced your question with a slightly different one since I suspect you have not written the question you intend, it's easy enough to let the index notation slip past at first glance. Let me know if you need further details, by the way you can use "@" with user name so I see you made a comment directed at me. – James S. Cook Jul 19 '13 at 2:48

Let me address the question "can the given tensor expression be contracted?" I would say no. Why? Well, consider:

1. $A_i A^i B^{ jrm} B_{jrm}$
2. $A_j A^i B^{ jrm} B_{irm}$
3. $A_i A^j B^{ irm} B_{jrm}$
4. $A_r A^i B^{ rmj} B_{imj}$

these cannot be further contracted since there is no free index in 1,2,3 or 4. The indices $i,j,r,m$ are all repeated hence there is already an implied summation and no new sums are possible.

Let me propose a slightly different question:

Question: given tensors $A$ and $B$ with ranks $1$ and $3$ respective, how can we construct new tensors by contraction? Note: we assume the existence of a metric $g_{ij}$ and inverse $g^{ij}$ for which the indices of $A$ and $B$ can be prepared covariantly or contravariantly. In particular, $A_i = g_{ij}A^j$ and $A^i = g^{ij}A_j$ and $B_{ijk} =g_{il}g_{jm}g_{kn}B^{lmn}$ or $B_{i}^{\ jk} = g_{il}B^{ljk}$ or $B^{i \ j}_{k} = g^{il}B_{lk}^{ \ \ j}$ and so forth...

With one copy of $A$ I can't see any contraction. However, with one copy of $B$ there are many different contractions possible (this is an incomplete list): $$B_{i \ j}^{\ i}, \qquad B_{ij \ }^{\ \ \ i},\qquad B_{j \ i}^{ \ \ i}, \qquad B^{i \ j}_{ \ i}, \qquad B^{j \ i}_{ \ i}$$ Now, moving on to two copies of $A$ we can construct: $$\underbrace{A_iA^i}_{\text{no free index, is scalar}}$$ Or, given one copy of $B$ and one copy of $A$ we could construct $$\underbrace{A_iB^{ijk}}_{\text{j,k free}}, \qquad \underbrace{A_iB^{ij}_{ \ \ j}}_{\text{nothing free, is a scalar}}$$ Let another possibility, suppose we have two copies of $A$ and one $B$ then contruct: $$\underbrace{A_iA_jB^{ijk}}_{\text{k free}}, \qquad \underbrace{A^iA^jB_{ijk}}_{\text{k free}}$$ Finally, if we have three copies of $A$ and one of $B$ then construct: $$\underbrace{A_iA_jA_kB^{ijk}}_{\text{no free index, a scalar}}, \qquad \underbrace{A^iA^jA^kB_{ijk}}_{\text{no free index, a scalar}}$$ Continuing this unending game, if we have two copies of $B$ and one of $A$ then construct: $$\underbrace{A_iB^{ijk}B_{jkl}}_{\text{free index is l}}, \qquad \underbrace{A_jB^{ijk}B_{ikl}}_{\text{free index is l}}, \qquad \underbrace{A_kB^{ijk}B_{ijl}}_{\text{free index is l}}$$ Generally speaking, if we have objects with ranks $r_1,r_2, \dots r_s$ and we contract say $k$-pairs of indices then the resulting object is a tensor of rank $r_1+r_2+ \cdots+ r_s-2k$. There are many different ways the pairs may be selected and any free index may be raised or lowered by the metric. The mathematically pure of heart are usually disgusted by these "ugly" formulas and would rather we understand these relationships in terms of isomorphisms which exist in the presence of a metric. If you desire such a treatment, I believe Barret O'neill's Semi-Riemannian geometry text constructs the formal mathematics behind these fun formulas.

• This is an excellent response!!! Thank you for taking the time to answer this one... I am very appreciative!! :) – CuriousGeorge119 Jul 19 '13 at 13:39