# Dimension of the space of algebraic Riemann curvature tensors

Given $n\in \mathbb N$, consider the vector space $\mathbb R^{n^4}$ whose elements I will denote by $(R_{abcd})$ with indices $a,b,c,d \in \{1, \dots, n\}$. This vector space is $n^4$-dimensional. The space of algebraic Riemann curvature tensors is the subspace $V$ of $\mathbb R^4$ consisting of all $(R_{abcd})$ satisfying the following symmetries:

• $R_{abcd} = - R_{bacd}$
• $R_{abcd} = - R_{abdc}$
• $R_{abcd} + R_{cabd} + R_{bcad} = 0$

How would one go about calculating $\dim V$?

Some thoughts: The three symmetries above imply another symmetry. Namely $R_{abcd} = R_{cdab}$. Combining this with the first two symmetries, it follows that $R_{abcd}$ can be viewed as a symmetric matrix of antisymmetric matrices. Since the space of antisymmetric matrices is $\binom n2$-dimensional, it follows that the first two symmetries together with the new one will cut out a space of dimension $$\frac{\binom n2 \left(\binom n2 +1\right)}{2}.$$ We still need to take into account the third symmetry. It is not quite clear to me which of the resulting additional equations are independent of each other and the old ones. But I believe there should certainly be some dependencies...

I noticed that if I only take into account those equations corresponding to values $a<b<c<d$ (of which there are $\binom n4$) and if I assume that these are all independent of each other, then I obtain that $$\dim V = \frac{\binom n2 \left(\binom n2 +1\right)}{2} - \binom n4 = \frac{n^2(n^2-1)}{12}.$$ (modulo miscalculation). This seems to be the right answer. However, I'm neither sure whether these equations indeed are independent nor am I sure whether they form a maximal set of independent equations.

I would appreciate your input! Thanks. =)

Its symmetries imply that an algebraic curvature tensor on the $$n$$-dimensional vector space $$V$$ can be viewed as an element of the space $$S^{2}(\wedge^{2}V^{\ast})$$ of symmetric bilinear forms on the second exterior power $$\wedge^{2}V^{\ast}$$, so the algebraic curvature tensors constitute a subspace of $$S^{2}(\wedge^{2}V^{\ast})$$.

An exterior four-form on $$V$$ can also be viewed as an element of $$S^{2}(\wedge^{2}V)$$, so $$\wedge^{4}V^{\ast}$$ is a subspace of $$S^{2}(\wedge^{2}V^{\ast})$$. Using the algebraic Bianchi identity, it is not hard to check that this subspace is the orthogonal complement of the subspace of curvature tensors. (Remembering this claim makes it easy to compute the dimension when one cannot remember it.)

Hence the dimension of the space of curvature tensors is \begin{align} \dim S^{2}(\wedge^{2}V^{\ast})& - \dim \wedge^{4}V^{\ast} = \frac{1}{2}\binom{n}{2}\left(\binom{n}{2} -1\right) - \binom{n}{4}\\ & = \frac{n(n-1)(n(n-1) - 2)}{8} - \frac{n(n-1)(n-2)(n-3)}{24}\\ & = \frac{n(n-1)\left(3n^{2} - 3n - 6 - n^{2} + 5n + 6\right)}{24}\\ & = \frac{n^{2}(n^{2}-1)}{12}. \end{align}

• +1 Or equivalently, this is the dimension of the module corresponding to the partition $(2,2)$ which is given by $A/B$ with $B=3\times 2\times 2\times 1$ is the product of hook lengths and $A=n\times (n+1)\times (n-1)\times n$ is the product of fillings of the boxes where one puts $n$'s on the diagonal and add (resp. subtract) $1$ when going to a higher (resp. lower) diagonal. Commented Feb 4, 2020 at 18:17

I saw that somewhere...

and that somewhere happens to be Weinberg, Gravitation and cosmology, around p.142-143, Description of curvature in N dimension. The calculation is made as you describe because Weinberg assumes in the following order (p.141):

1. $R_{abcd}=R_{cdab}$

2. $R_{abcd}= -R_{bacd}=-R_{abdc}$

3. $R_{abcd}+ R_{bcad}+ R_{cabd}=0$

The first and third are indeed redundant

$3. \Rightarrow 1.$ : (so one can actually discard 1.) Consider $R_{abcd}-R_{cdab} \stackrel{?}{=}0$.

Replace both term with 3. ($R_{abcd}= -R_{bcad}- R_{cabd}$ same for the other) and use antisymmetry to cancel two terms. The equality holds for all $a,b,c,d$ so write the obtained equation again, substituting $"a":=b, "b":=c, "c":=d , "d":=a$. Cancel some terms.

Conversely, if two of the four indexes $a,b,c,d$ coincide, then $1. \Rightarrow 3.$ (assuming 2.) by a direct check.

So 3. only adds extra independent constraints if the four indexes are different.

ps: - i see no miscalculation. - of course the extra ${n\choose 4}$ equations 3. are independent since for $\{a_1,b_1,c_1,d_1 \}\neq \{a_2,b_2,c_2,d_2 \}$ no permutation can map a list of elements of the first set to a list of elements of the second set