You mean at most 2.3727. It's not really a safe assumption that the bound will shrink asymptotically to the real omega as far as I know. But it is suspected that higher tensor powers may drop the exponent slightly.
I believe there are approaches related to other constructions in group theory, such as the wreath product, which may eventually lead to better results. It's just a huge unknown at this point. (My knowledge of these techniques can be written on the back of a postage stamp otherwise I'd try and provide more detail.)
One certainly hopes that higher and higher tensor powers are not the way to get omega = 2, as the algorithms just become more and more complex.
Someone may correct me if I am wrong but I think it is still not known even what the best algorithm is which breaks matrices up into 3x3 and whether one exists with better complexity than Strassen even (which breaks the matrix up into 2x2). A huge number of CPU cycles have been used trying to find the best algorithm using a 3x3 decomposition.
This (http://www.siam.org/pdf/news/174.pdf) short, highly readable paper gives a high-level overview of the group-theoretic approach. Here are the key points, quoted from the paper:
* "What Cohn and Umans realized is that under certain conditions, matrix multiplication, too, can be embedded into the group algebra of a finite
group G, although the embedding is more complex and the group must be non-abelian to yield a nontrivial bound."
The conditions that must be satisfied are as follows:
* "...the group has three subgroups, H_1, H_2, H_3, with the property that whenever h_1 ∈ H_1, h_2 ∈ H_2, and h_3 ∈ H_3 with h_1 h_2 h_3 = 1, then h_1 = h_2 = h_3 = 1" (AKA the "triple-product property,") and
* "|H1||H2||H3| > Σd_i^3," where the d_i's are the dimensions of the square submatrices into which we decompose the original matrix. (AKA "beating the sum of cubes.")
The groups that have been found to have these properties are "wreath products of abelian groups with the symmetric group S_n."
Sorry, I got confused there for a minute. The answer is in fact just "yes". I'm not sure why this was voted down.
The reason is that the border rank of 2x2 matrix multiplication is seven. This was proved 5 years ago by Landsberg, though it took quite some effort for me to track this down. Sometimes "yes" is the best one can do at short notice. Perhaps individuals who enjoy taking points away from people might consider this!
However, Winograd's algorithm can do nxn matrix multiplication with n^3/2 + 2n^2 multiplications (and many more additions than usual). This is only useful if the coefficients are very large because of the cost of the extra additions. Moreover, this is not a block algorithm, i.e. it cannot be applied recursively. Anyhow, for 2x2 matrices, n = 2 and so the total number of multiplications is 2^3/2 + 2n^2 = 12 which is more than the usual 8 multiplications (or 7 if you use Strassen). Similarly for n = 3 it is less efficient than the naive 27 multiplications (and even less efficient than Lederman's algorithm: 23 multiplications and Bard's algorithm: 22 multiplications), but for n = 4 we get 4^3/2 + 2x4^2 = 64 which is the same as the naive algorithm and for n = 6 we get 6^3/2 + 2x6^2 = 180 which is well under the usual 216 multiplications required by the naive algorithm and even less than the number of multiplications required if you use one level of Strassen. Already by n = 8 Strassen is better if you go by the number of multiplications alone. Of course in practice the coefficients need to be really large (perhaps tens of machine words) before either Winograd or Strassen are better than the naive algorithm.
Note that if you break a larger matrix up into 2x2, i.e. 4 blocks then the smaller submatrices no longer commute (matrix multiplication is not commutative). This is why Winograd's algorithm is not a block algorithm like the Strassen method.
I believe there are approaches related to other constructions in group theory, such as the wreath product, which may eventually lead to better results. It's just a huge unknown at this point. (My knowledge of these techniques can be written on the back of a postage stamp otherwise I'd try and provide more detail.)
One certainly hopes that higher and higher tensor powers are not the way to get omega = 2, as the algorithms just become more and more complex.
Someone may correct me if I am wrong but I think it is still not known even what the best algorithm is which breaks matrices up into 3x3 and whether one exists with better complexity than Strassen even (which breaks the matrix up into 2x2). A huge number of CPU cycles have been used trying to find the best algorithm using a 3x3 decomposition.