3
$\begingroup$

The question in my assignment: Suppose we have a tensor $A^{\mu\nu\alpha\beta}$ in four spacetime dimensions. This tensor is antisymmetric in the first two indices, i.e., $A^{\mu\nu\alpha\beta}=-A^{\nu\mu\alpha\beta}$ and symmetric in last two indices, i.e., $A^{\mu\nu\alpha\beta}=A^{\mu\nu\beta\alpha}$. Determine the number of independent components this tensor has. On the other hand, if the tensor is antisymmetric in all four indices how many independent components it will have? In general, if we have a '$n$' dimensions, how many independent components it will have.

My answer: As the tensor $A^{\mu\nu\alpha\beta}$ is anti-symmetric under exchange of its first two indices, there are $\frac{4(4-1)}{2}=6$ independent combinations for $\mu$ and $\nu$. Now, for each of these $6$ combinations there are $\frac{4(4+1)}{2}=10$ independent combinations of $\alpha$ and $\beta$, as the tensor is symmetric under the exchange of these two indices. Thus, there are in total $6\times 10=60$ independent components of the tensor.

If the tensor is anti-symmetric in all its four indices, then:\par As the indices cannot be repeated, thus the first index has $4$ numbers to choose from; once that is done for the second index we have only $3$ choices; for the third index $2$ choices and the last index is determined. The number of possible combinations is $4\times3\times2=4!$. But all these combinations can be obtained from permuting a single combination, as there are $4!$ possible permutations, therefore, the number of independent components is $\frac{4!}{4!}=1$

Number of independent components for a fully antisymmetric $(4,0)$ rank tensor in $n$ dimension:\par As the indices cannot be repeated, thus the first index has $n$ numbers to choose from; once that is done for the second index we have only $n-1$ choices; for the third index $n-2$ choices and the last index has $n-3$ choices. Therefore, the number of possible combinations $n\times(n-1)\times(n-2)\times(n-3)=\frac{n!}{(n-4)!}$. Again due to the total antisymmetry, once one combination of indices is determined, the rest can be obtained by permutations. As there are $4!$ possible permutations, the number of independent components $\frac{n!}{4!(n-4)!}={}^nC_4$.

Question: (1) Whether my arguments are correct.

(2) Is there a list for most general formulas for calculating independent components of tensors in various situations? Or maybe someone can list a few with explanations.

$\endgroup$
1
+50
$\begingroup$

Note that we expect there to be $n^4$ components to start out with for an arbitrary $(4,0)$ tensor $T^{abcd}$ in $n$ dimensions. (and in general a generic $(m,0)$ tensor in $n$ dimensions should have $n^{m}$ components)

(a) Start with the antisymmetric case where $A^{abcd} = - A^{bacd}$. Notice that for any $a=b$ we end up having $A^{aacd} =0$, which is sort of like having a $(3,0)$ tensor with all components zero. This means that you would expect $n^3$ components to be zero, so at this point there are $n^4 - n^3$ components left. We also note that for $a \neq b$ we also always have $T^{bacd} = - T^{abcd}$, which implies that half the remaining components are independent: this means there are a total of $\frac{1}{2} \cdot (n^4 - n^3) = \frac{n(n-1)}{2} \cdot n^2$ free components for an antisymmetric tensor of this form.

(b) For the symmetric case $S^{abcd} = S^{abdc}$, the argument is similar, except your 'diagonals' are now free components. As in the above (but now $S^{abdc} = S^{abcd}$ for $c \neq d$), there are $\frac{1}{2} \times (n^4 - n^3)$ free components which are 'off-diagonal', and so now just add to this the extra $n^3$ free diagonal components $S^{abcc}$. The total is $\frac{1}{2} \cdot (n^4 - n^3) + n^3 = n^2 \cdot \frac{n(n+1)}{2}$

(c) If you have a tensor with both properties (b) and (c), the arguments in the above follow through similarly (because the symmetries act on separate sets of indices), and can be phrased as you did --- the first two indices being anti-symmetric mean there are $\frac{n(n-1)}{2}$ free combinations of $a$ and $b$, and the latter two indices have $\frac{n(n+1)}{2}$ free combinations. Overall the tensor has $\frac{n(n-1)}{2} \cdot \frac{n(n+1)}{2} = \frac{n^2 (n-1)(n+1)}{4}$ free components. That is equal to $60$ for $n=4$.

(d) Finally for the tensor $F^{abcd}$ which is antisymmetric in all its indices (also known as a completely/totally antisymmetric tensor). You've got the right answer and the arguement is correct. Interestingly in $n = 4$ dimensions, having 1 free component means that the only type of totally antisymmetric $(4,0)$ tensor you can have is proportional to the Levi-Cevita tensor (and this is generically true for a totally antisymmetric $(m,0)$ tensor in $n$ dimensions for $n=m$).

$\endgroup$
4
  • $\begingroup$ In the answer (a) you're only considering the antisymmetric property, however the question clearly states that the tensor is anti-symmetric in first two indices but symmetric in last two. So why haven't you taken this property into account? $\endgroup$ – Faber Bosch Oct 22 '20 at 19:49
  • 1
    $\begingroup$ My bad, I misread the question (thought you were asking for each case separately). I added in some content to answer $\endgroup$ – QuantumEyedea Oct 23 '20 at 1:44
  • $\begingroup$ Okay! Thanks! Your answer looks good. But only one thing, if happen you know any source that discusses this types of problems a little longer than the usual GR books, please mention that. I'll wait for two more days, and if by then no other answer shows more examples, I'll reward the bounty. $\endgroup$ – Faber Bosch Oct 23 '20 at 3:49
  • 1
    $\begingroup$ I'm not sure of a resource which goes into great length about that. I can say that Caroll's Spacetime and Geometry has a good and explicit discussion in Chapter 3.7 on the number of free components of the Riemann tensor, which might be useful for you. $\endgroup$ – QuantumEyedea Oct 23 '20 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.