Strassen $2\times 2$ Matrix Multiplication from a 3-dimensional Volume Form

Strassen’s $2\times 2$ matrix multiplication algorithm [1] is a formula for multiplying $2\times 2$ matrices $a$ and $b$:

where $I$ is the identity matrix, $\mathrm{tr}$ is the trace, and the $X_{i},Y_{i},Z_{i}$ are constant matrices. This formula is a rank 7 decomposition of the matrix multiplication tensor, that is, a decomposition of matrix multiplication into a sum of 7 simple tensors.

This may be applied recursively to multiply $n\times n$ matrices in $O(n^{\log 7/\log 2})$ time, approximately $O(n^{2.81})$, opening a research field to which the book [2] provides an introduction. One line of research has focused on further improving this asymptotic complexity, notably [3], [4] achieving $O(n^{2.376})$ and several refinements, recently [5] and [6], approaching $O(n^{2.37})$. Another line of research has pursued decompositions of matrix multiplication tensors for other small matrix sizes such as $3\times 3$, $4\times 4$, etc. These have often involved numerical searches, such as the recent [7].

Despite these advances, matrix multiplication algorithms faster than $O(n^{2.81})$ are “almost never implemented” [8], and practical evaluations such as [9] have continued favoring the Strassen $2\times 2$ algorithm. Known algorithms for other small matrix sizes struggle to significantly improve on it: the up-to-date table [10] shows complexity exponents clustering around 2.8. For $4\times 4$ matrix multiplication, the Strassen algorithm has tensor rank $7^{2}=49$, and that remained the state of the art for over 55 years until [7] and [11] lowered that from 49 to 48, achieving a complexity exponent of $2.79$. Moreover, known algorithms with substantially lower asymptotic complexity tend to have large constants in the $O$, as discussed in [12].

The Strassen $2\times 2$ algorithm also stands out from a theoretical perspective: its tensor rank 7 is known to be optimal [13] and it is known to be essentially unique under that constraint [14]. By contrast, the tensor rank of $n\times n$ matrix multiplication is still unknown for all $n\geq 3$. For $n=3$, it is still only known to be between 19 and 23, see [15].

Optimality and uniqueness make the Strassen $2\times 2$ algorithm a basic fact of 2-dimensional linear algebra. Such facts are expected to be simple and geometric. However, the original statement and proof of the Strassen algorithm are calculations on matrix coefficients. This has motivated a quest for geometric interpretations. The recent [16] and [8] in particular were inspirational to the present article, and [8] contains a survey of this endeavour, tracing it back to the years following the publication of the original Strassen article [1]. Other recent articles in this line of research include [17], [18], [19] and [20].

The present article offers a geometric interpretation of the Strassen algorithm by addressing a more general question: is the Strassen algorithm an independent fact in multilinear algebra, or could it be related to a known fact? We derive it from the expansion of a 3-dimensional volume form into an antisymmetrized sum of $3!=6$ simple tensors. That expansion follows from the one-dimensionality of the space of antisymmetric $n$-forms, which is an abstract version of Cavalieri’s principle, the idea that the volume of a solid is unchanged by sliding parallel slices. As to the question of why specifically $2\times 2$ matrices, the answer is that as matrix multiplication is a tensor of order 3 on matrix spaces, interpreting it as a volume form requires a 3-dimensional matrix space, and the specific case of $2\times 2$ matrices gives us such a 3-dimensional matrix space by taking the quotient by multiples of the identity matrix: $3=2^{2}-1$.

Acknowledgements. The author would like to thank Paolo d’Alberto and Zach Garvey at AMD for helpful comments.

Fix, for this entire article, a 2-dimensional vector space $V$ over a field $k$. Let $\mathrm{L}(V)$ denote the space of linear maps from $V$ to itself. Start by considering this trilinear form $g$ on $\mathrm{L}(V)$:

We notice (Lemma 12) that $g$ is a volume form on the quotient of $\mathrm{L}(V)$ by the multiples of the identity matrix, which has dimension 3. This gives (Lemma 13) rank 6 decompositions of $g$ parametrized by bases of the dual space. Our next step is to relate $g$ to this other trilinear form $h$ on $\mathrm{L}(V)$:

Using the natural isomorphism $\mathrm{L}(V)^{*\otimes 3}\simeq\mathrm{L}(V^{\otimes 3})$, view the trilinear forms $\mathrm{tr}(a_{1})\mathrm{tr}(a_{2})\mathrm{tr}(a_{3})$, $\mathrm{tr}(a_{1}a_{2}a_{3})$ and $\mathrm{tr}(a_{3}a_{2}a_{1})$ as respectively the permutations $\mathrm{id}$, (123) and (321) permuting the terms in $V^{\otimes 3}$ (Lemma 8). This allows viewing $h$ as the composition of $g$ with a linear map induced by the permutation (321) (Lemma 16), which allows transporting certain rank 6 decompositions of $g$ into rank 6 decompositions of $h$ (Proposition 18, our main result), yielding (Corollary 20)

which is a rank 7 decomposition of $\mathrm{tr}(a_{1}a_{2}a_{3})$. Dualizing that yields a rank 7 decomposition of matrix multiplication (Corollary 21) parametrized by a choice of basis. A specific choice yields the original Strassen algorithm (Corollary 22).

Throughout this article, “vector space” means finite-dimensional vector space. For any vector spaces $U$ and $W$ over a field $k$, let $\mathrm{L}(U,W)$ denote the space of linear maps from $U$ to $W$. In the case $W=U$, we write $\mathrm{L}(U)$ for $\mathrm{L}(U,U)$. In the case $W=k$, we let $U^{*}=\mathrm{L}(U,k)$ denote the dual space of $U$.

Given any vector spaces $U_{1},\ldots,U_{n}$, $W_{1},\ldots,W_{n}$, we will make the identification

As special cases of that, for any vector space $U$ over $k$, for any positive integer $n$, we identify $\mathrm{L}(U)^{\otimes n}\simeq\mathrm{L}(U^{\otimes n})$ and $U^{*\otimes n}\simeq(U^{\otimes n})^{*}$. The latter identification means concretely that given linear forms $\mu_{1},\ldots,\mu_{n}$ on a vector space $U$, we identify the tensor $\mu_{1}\otimes\ldots\otimes\mu_{n}$ as the $n$-linear form on $U$ given, for all vectors $u_{1},\ldots,u_{n}$ in $U$, by:

Let us now describe a few other natural isomorphisms of tensor spaces that we will keep as named isomorphisms, refraining from making identifications.

The following lemma is classical and is encountered around Weyl invariant tensor theory ([21], [22]), which, among other things, establishes that the $t_{\sigma}$ span the space of $\mathrm{GL}(U)$-invariant tensors in $\mathrm{L}(U^{\otimes 3})$. We will not need any of that theory, but it is still useful context.

In the next section, in the proof of our main result (Proposition 18), we will need the following Lemma 10, which is about composing the $L_{t_{\sigma}}$ with forms that are simple tensors. In the proof of Lemma 10, we will need this simpler lemma about evaluating the $L_{t_{\sigma}}$ on simple tensors:

Let us return to the 2-dimensional vector space $V$ over a field $k$ that we had fixed in the overview. Let $I$ denote the identity in $\mathrm{L}(V)$.

Note that $\dim Q=(\dim V)^{2}-1=3$. The dual $Q^{*}$ is identified with the subspace of $\mathrm{L}(V)^{*}$ consisting of those forms $\mu$ that satisfy $\mu(I)=0$.

While Lemma 13 allowed arbitrary linear forms $\mu_{i}$, Proposition 18 will need to restrict to rank one forms, meaning the $\iota^{*}(v\otimes\lambda)$ for $v$ in $V$ and $\lambda$ in $V^{*}$. The necessity of that restriction is discussed in Remark 19.

To elaborate on the previous remark, the linear form $a\mapsto\mathrm{tr}(a)$ does not have rank one, so even though $q$ has rank one as a tensor of order 3 in $\mathrm{L}(V)^{*\otimes 3}$, it does not have rank one as a tensor of order 6 in $(V\otimes V^{*})^{\otimes 3}$, and our tool for transporting tensor decompositions, Lemma 10, applies to tensors of order 6 in $(V\otimes V^{*})^{\otimes 3}$.

Proposition 18 is already a form of Strassen’s algorithm, but that may be obscured by the tensor formalism, so let us derive a few more concrete statements as corollaries.