Algebraic Invariants of Tensors: Algorithms and Decompositions

Chris Liu

Dissertation Defense

May 8, 2026

Plan

Tensors and their algebraic invariants
First result: a faster algorithm to compute the adjoint and derivation algebra of tensors
Second result: theorem on the algebraic invariants on the product of tensors

Tensors and their algebraic invariants

$ T $ is $ (a \times b \times c) $ grid of numbers

Tensors are multiway grids of numbers with a multilinear interpretation

As a bilinear map

\[ t: \mathbb{R}^a \times \mathbb{R}^b \rightarrowtail \mathbb{R}^c \]

As a trilinear form

\[ \tau: \mathbb{R}^a \times \mathbb{R}^b \times \mathbb{R}^c \rightarrowtail \mathbb{R}\]

\[\tau \left(\sum_i u_i e_i, \sum_j u_je_j, \sum_k u_ke_k \right) = \sum_{ijk} T_{ijk} u_iv_jw_k\]

( $ \rightarrowtail $ for multilinear )

Example 1: Discord chat logs

A cube of numbers, where $ (i,j,k) $ is the number of times word $i$ was said by user $j$ at time $k$

$ \text{Users} = \{ \text{Andrea}, \text{Bob}, \text{Chris}, \text{Dave}, \text{Eve}, \ldots \} $

$ \text{Days of the week} = \{\text{Monday}, \text{Tuesday}, \ldots \} $

$ \text{Words} = \{\text{apple}, \text{break}, \ldots,\text{goalie}, \text{hockey}, \ldots \} $

$ \text{Groups} = \{ \alpha_1 \cdot \text{Chris} + \alpha_2 \cdot \text{Dave} + \alpha_3 \cdot \text{Eve}, \ldots \} $

$ \text{Time clusters} = \{\beta_1 \cdot \text{Saturday} + \beta_2 \cdot \text{Sunday}, \ldots \} $

$ \text{Themes} = \{ \gamma_1\cdot \text{hockey} + \gamma_2 \cdot \text{puck} + \gamma_3 \cdot \text{goalie}, \ldots \} $

Linear combinations

Data mining algorithm

Example 2: Multiplication

Let $ A $ be a $\mathbb{F}$-vector space with bilinear multiplication $ \mu: A \times A \rightarrowtail A $ ($\mathbb{F}$-algebra)

For ordered basis $ (e_1,\ldots, e_n )$ of $A$, coordinatize $\mu$ by a $ (n \times n \times n) $ grid of numbers $T$ satisfying $ \mu(e_i,e_j) = \sum_k T_{ijk}e_k $.

For $A = \frac{\mathbb{F}[x]}{x^2-3x+2} $

Change of bases

M = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}

Algebraic invariants detect structure

Source: Optimal Search Spaces for Tensor Problems, Figure 1.1

Adjoints detect

Given $ t: U \times V \rightarrowtail W $

$$\operatorname{Adj}(t) = \{(X,Y) \mid (\forall u,v) \;\; t(Xu,v) = t(u,Yv) \}$$

Analogy: Given a bilinear form

$ \langle \cdot \mid \cdot \rangle : U \times U \rightarrowtail \mathbb{F} $ and $A \in \operatorname{End}(U)$, its adjoint $ A^{\ast} $ satisfies $ \langle Au \mid v \rangle = \langle u \mid A^{\ast} v \rangle $

(transpose for the dot product)

Myasnikov '90, Meataxe (Parker-Norton '84, Ronyai '89), Wilson '08 and others use the algebra, i.e

Theorem

Exists $\mathcal{E} = \{e_1,\ldots, e_n \} \subset \text{Adj}(t) $,

$ \sum_i e_i = 1$, and $e_ie_j = e_i $ if $ i = j $, otherwise $0$

if and only if

Exists $\perp$-decomposition, $U := \bigoplus_i U_i$ and $V := \bigoplus_i V_i $,

\[ t(U_i, V_j) = t(U_i,V_i) \quad \text{if } i = j, \text{otherwise } 0 \]

Derivations detect

For $ (u,v,w) $ eigenvectors of $ (X,Y,Z) $ in $ \text{Der}(\tau) $ with eigenvalues $ (\kappa , \lambda, \rho) $,

By distributive property,

\[ (\rho - \kappa - \lambda) (\tau(u,v,w)) = 0 \]

Source: Detecting cluster patterns in tensor data, Figure 3

Given $ t: U \times V \rightarrowtail W $

$$\operatorname{Der}(t) = \{(X,Y,Z) \mid (\forall u,v) \; Z(t(u,v)) = t(Xu,v) + t(u,Yv) \}$$

Analogy: The product rule in Calculus to understand multiplication. The derivative $ \frac{d}{dx}(fg) = (\frac{d}{dx}f)g + f(\frac{d}{dx}g) $

The surface $z=xy$

Eigenbasis gives cluster patterns

\[\tau(u,v,Zw) = \tau(Xu,v,w) + \tau(u,Yv,w)\]

\[ \tau(u,v,\rho w))-\tau(\kappa u,v,w) - \tau(u, \lambda v,w) = 0\]

Theorem (Brooksbank, Kassabov, Wilson '24)

(For trilinear form $\tau: U \times V \times W \rightarrowtail \mathbb{F} $ )

Tensors vs tensor product of vector spaces?

Given $U,V$ vector spaces, the tensor product space is $ (U \otimes V, \varphi: U \times V \rightarrowtail U \otimes V) $

such that for all bilinear maps $f: U \times V \rightarrowtail W$, there exists unique induced linear map $\tilde{f}: U \otimes V \rightarrow W$ such that $f(u,v) = \tilde{f}(\varphi(u,v))$.

$U \otimes V$ is spanned by $ \{ \varphi(u,v) : u \in U, v \in V \}$
Pure tensors are elements $ \varphi(u,v) \in U \otimes V $, often denoted as $u \otimes v$
Elements of $U \otimes V$ are tensors in our sense after identifying $ U \otimes V $ with $(U \otimes V)^{\ast} $ for instance by a basis

"Theorem" (L.)

Faster (practical and theoretical) algorithm to compute the adjoint and derivation algebra of tensors

Three step algorithm

solving a smaller restricted subsystem
lifting the subsystem solution to a global candidate solution
verify the global candidate solution

Let $A \in \mathbb{F}^{a \times b \times c} $

$ (R,S,T) = (A,-A, 0) $ gives adjoint

(Joint with James B. Wilson, Joshua Maglione)

R \in \mathbb{F}^{r \times b \times c}\\ S \in \mathbb{F}^{a \times s \times c}\\ T \in \mathbb{F}^{a \times b \times c}

X \in \mathbb{F}^{a \times r}\\ Y \in \mathbb{F}^{s \times b}

Given

Find

Such that

(\forall i \in [c]) \; XR_i + S_iY = T_i

Simultaneous Sylvester System

Recall given $ t: \mathbb{F}^{a} \times \mathbb{F}^{b} \rightarrowtail \mathbb{F}^{c} $ $$\operatorname{Adj}(t) = \{(X,Y) \mid (\forall u,v)\;\; t(Xu,v) = t(u,Yv)\}$$

Corresponding baseline linear system

Example

R_1 = \begin{bmatrix} 1 & 5 \\ 0 & -1 \\ 2 & 11 \end{bmatrix}, \quad R_2 = \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ -1 & 1 \end{bmatrix}

S_1 = \begin{bmatrix} 1 & 0 & 3 & 0\\ 1 & 1 & 10 & -1 \end{bmatrix}, \quad S_2 = \begin{bmatrix} 0 & 1 & 7 & -1\\ 1 & -1 & -4 & 1 \end{bmatrix}

T_1 = \begin{bmatrix} 1 & 19\\ -11 & 4 \end{bmatrix}, \quad T_2 = \begin{bmatrix} -10 & 4\\ 8 & 2 \end{bmatrix}

\begin{array}{|ccc|ccc|cccc|cccc||c|} \hline X_{11} & X_{12} & X_{13} & X_{21} & X_{22} & X_{23} & Y_{11} & Y_{21} & Y_{31} & Y_{41} & Y_{12} & Y_{22} & Y_{32} & Y_{42} & T_{ijk} \\ \hline 1 & 0 & 2 & & & & 1 & 0 & 3 & 0 & & & & & 1 \\ 0 & 1 & -1 & & & & 0 & 1 & 7 & -1 & & & & & -10 \\ \hline 5 & -1 & 11 & & & & & & & & 1 & 0 & 3 & 0 & 19 \\ 1 & 1 & 1 & & & & & & & & 0 & 1 & 7 & -1 & 4 \\ \hline & & & 1 & 0 & 2 & 1 & 1 & 10 & -1 & & & & & -11 \\ & & & 0 & 1 & -1 & 1 & -1 & -4 & 1 & & & & & 8 \\ \hline & & & 5 & -1 & 11 & & & & & 1 & 1 & 10 & -1 & 4 \\ & & & 1 & 1 & 1 & & & & & 1 & -1 & -4 & 1 & 2\\ \hline \end{array}

(LMW) Idea: Avoid big linear system altogether with a solve-lift-check approach

Linear system given by matrix of size $ (abc) \times (ar + bs) $ is $O(n^7) $ to solve using Gaussian Elimination

($n = a+b+c+r+s $)

\text{Recall } R \in \mathbb{F}^{r \times b \times c}, S \in \mathbb{F}^{a \times s \times c}, T \in \mathbb{F}^{a \times b \times t}

Permute rows & columns

Idea: solve-lift-check approach

$ I $

$ J $

$ M_R $

$ M_S $

(L.) Distill to

Solve $ Mx=b $ systems
Solve $ MX=N(\theta) $ parametrized systems
Solve $Au+\alpha = Bv + \beta$ systems

Restrict to $I \subset [a]$ for $S$, and $J \subset [b] $ for $R$
Solve restricted system for $ (X_{DR}, Y_{DR}) $
Backsub $ M_R X = N_R(Y_{DR}) $
Backsub $ M_S Y = N_S(X_{DR}) $
Verify

Given $M \in \mathbb{F}^{m \times n} $, and $ N: \Theta \rightarrow \mathbb{F}^{m \times \ell} $ affine map, solve for $X \in \mathbb{F}^{n \times \ell} $

Given matrices $A,B$ and constants $ \alpha, \beta$, solve for all pairs $ (u,v) $

regular family of instances

(Bounded dimension) $ \operatorname{dim}(\Theta_{DR}) \leq C$
(Lift feasibility if restricted system has solutions)
- $ N_R(Y_{DR}) \subseteq \operatorname{Im}(M_R) $
- $ N_S(X_{DR}) \subseteq \operatorname{Im}(M_S) $
(Lift uniqueness) $ \operatorname{Ker}(M_R) = \operatorname{Ker}(M_S) = \{0\} $

Theorem A (L.)

Given a regular family, and a black-box algorithm for $ (I,J) $, the solve-and-lift approach takes $O(n^4)$ arithmetic operations for deterministic solutions.

Remark - The black-box restriction algorithm needed for determinstic runtimes. $ \exists $ randomized algorithms.

$ I $

$ J $

$ M_R $

$ M_S $

Recall backsub $ M_R X = N_R(Y_{DR}) $, $ M_S Y = N_S(X_{DR}) $

Assumption is reasonable for natural classes of instances (e.g generic cubic tensors with overdetermined restricted system)

Derivation Equation

Let $A \in \mathbb{F}^{a \times b \times c} $

$ (R,S,T) = (A,A, A) $ gives derivation

Recall given $ t: \mathbb{F}^{a} \times \mathbb{F}^{b} \rightarrowtail \mathbb{F}^c $

$$\operatorname{Der}(t) = \{(X,Y,Z) \mid (\forall u,v) \; Z(t(u,v)) = t(Xu,v) + t(u,Yv) \}$$

R \in \mathbb{F}^{r \times b \times c}\\ S \in \mathbb{F}^{a \times s \times c}\\ T \in \mathbb{F}^{a \times b \times t}

X \in \mathbb{F}^{a \times r}\\ Y \in \mathbb{F}^{s \times b}\\ Z \in \mathbb{F}^{t \times c}

Given

Find

Such that

(\forall i \in [c]) \; XR_i + S_iY = \sum_j T_jZ_{ji}

Can you tell $R \neq S \neq T $ here?

Theorem B (L.)

Given a regular family, and a black-box algorithm for $ (I,J,K) $, the solve-and-lift approach takes $O(n^{4.5})$ arithmetic operations for deterministic solutions.

Analogous derivation regular family ensure the bounded dimension, and unique lifting property

Restrict to $I \subset [a]$ for $S$, $J \subset [b] $ for $R$, and $K \subset [c] $ for $T$
Solve restricted system for $ (X_{TR}, Y_{TR}, Z_{TR}) $
Backsub for $ X,Y,Z $
Verify on unrestricted indices

Remark - Extra $ n^{0.5} $ factor is due to the restricted system having $ O(n^{1.5})$ variables

t = [t_1, t_2], \text{ with} \\ t_1 = \begin{bmatrix} 10 & 5\\ 5 & 6 \end{bmatrix} \quad t_2 = \begin{bmatrix} 9 & 3\\ 3 & 7 \end{bmatrix}

\begin{array}{|cccc|cccc|cccc|} \hline x_{11}&x_{12}&x_{21}&x_{22} &y_{11}&y_{21}&y_{12}&y_{22} &z_{11}&z_{12}&z_{21}&z_{22} \\\hline 10& 5& & & 10& 5& & & -10& & -9& \\ 9& 3& & & 9& 3& & & & -10& & -9\\ 5& 6& & & & &10& 5 & -5& & -3& \\ 3& 7& & & & & 9& 3 & & -5& & -3\\ & &10& 5 & 5& 6& & 0 & -5& & -3& \\ & & 9& 3 & 3& 7& & 0 & & -5& & -3\\ & & 5& 6 & & & 5& 6 & -6& & -7& \\ & & 3& 7 & & & 3& 7 & & -6& & -7\\ \hline \end{array}

(baseline) $ \text{Der}(t) $ is nullspace of

source: https://github.com/cliu587/fast-der-solver

Performance on regular family with one solution realizes theoretical improvements.

Performance on non-regular family with $n$ solutions for $ (n \times n \times n) $ tensors still shows improvement

Extras cut for time

Remark - As $O(n^4)$ cost is needed to determinstically verify a solution, this is the best complexity for a determinstic solution we can expect.

"Theorem" (L.)

Decomposing the algebraic invariants of the product of tensors

Tensor products of algebras

Let $ A, B $ be unital associative $ \mathbb{F} $-algebras

$ A \otimes B $ is a unital associative $\mathbb{F}$-algebra, with multiplication
\[ (a \otimes b)(c \otimes d) = ac \otimes bd \]

Example: Kronecker product of matricies

Example: Extending scalars (Recall $ \mathbb{H} $ are the Quaternions)

$ u = i_{\mathbb{H}} \otimes i_{\mathbb{C}} $ satisfies $ u^2 = 1 $, so the idempotent $\frac{1+u}{2} $ splits the algebra

\mathbb{H} \otimes \mathbb{C} \cong \mathbb{M}_2(\mathbb{C})

\[ 1 \otimes 1 \mapsto \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, i \otimes 1 \mapsto \begin{bmatrix}i & 0\\ 0 & -i\end{bmatrix}, j \otimes 1 \mapsto \begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}, k \otimes 1 \mapsto \begin{bmatrix}0 & i \\ i & 0\end{bmatrix}\]

Define on multiplication tables

\begin{array}{|c|cc|} \hline & 1 & x \\ \hline 1 & 1 & x \\ x & x & -1 \\ \hline \end{array} \otimes \begin{array}{|c|cc|} \hline & 1 & y \\ \hline 1 & 1 & y \\ y & y & -5y-10 \\ \hline \end{array} = \begin{array}{|c|cccc|} \hline & 1 & x & y & xy \\ \hline 1 & 1 & x & y & xy \\ x & x & -1 & xy & -y \\ y & y & xy & -5y-10 & -5xy-10x \\ xy & xy & -y & -5xy-10x & 5y+10 \\ \hline \end{array}

$\frac{\mathbb{F}[y]}{y^2+5y+10} $

$ \mathbb{F}[\sqrt{-1}] $

Extend to heterogeneous maps $ s: U \times V \rightarrowtail W $ and $t: X \times Y \rightarrowtail Z $

Let $ s: U \times V \rightarrowtail W $ and $t: X \times Y \rightarrowtail Z $.

Define $ s \otimes t: (U \otimes X) \times (V \otimes Y) \rightarrowtail (W \otimes Z) $ as
\[ (s \otimes t)(u\otimes x, v \otimes y) = s(u,v) \otimes t(x,y) \]

Product of bilinear maps

Let $s: \mathbb{F}^2 \times \mathbb{F} \rightarrowtail \mathbb{F}^2 $ and $ t: \mathbb{F} \times \mathbb{F}^3 \rightarrowtail \mathbb{F}^3 $ both be bilinear maps corresponding to scaling by $ \mathbb{F} $

$ s \otimes t $ is a bilinear map from $ (\mathbb{F}^2 \otimes \mathbb{F}) \times (\mathbb{F} \otimes \mathbb{F}^3)$ to $(\mathbb{F}^2 \otimes \mathbb{F}^3) $

$ s \otimes t \cong r $, for $r: \mathbb{F}^2 \times \mathbb{F}^3 \rightarrowtail \mathbb{F}^6 $ the outer product tensor

Example

(s \otimes t) \left( \begin{bmatrix}u_1 \\ u_2 \end{bmatrix} \otimes c_1, c_2 \otimes \begin{bmatrix}v_1 \\ v_2 \\ v_3 \end{bmatrix} \right) = s \left( \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}, c_2 \right) \otimes t \left( c_1, \begin{bmatrix}v_1 \\ v_2 \\ v_3 \end{bmatrix} \right)

= c_1c_2 \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \otimes \begin{bmatrix}v_1 & v_2 & v_3 \end{bmatrix} = c_1c_2 \begin{bmatrix} u_1v_1 & u_1v_2 & u_1v_3 \\ u_2v_1 & u_2v_2 & u_2v_3 \end{bmatrix}

Axis action equalities

Let $t: U_2 \times U_1 \rightarrowtail U_0$. For $\alpha \in \operatorname{End}(U_2), \beta \in \operatorname{End}(U_1), \gamma \in \operatorname{End}(U_0)$, axis actions are

$$\mathcal{L}(t) = \{(\alpha, \gamma) : t \bullet_2 \alpha = t \bullet_0 \gamma \}$$

$$\mathcal{M}(t) = \{(\alpha, \beta) : t \bullet_2 \alpha = t \bullet_1 \beta \}$$

$$\mathcal{R}(t) = \{(\beta, \gamma) : t \bullet_1 \beta = t \bullet_0 \gamma \}$$

$$\mathcal{C}(t) = \{(\alpha, \beta, \gamma) : t \bullet_2 \alpha = t \bullet_1 \beta = t \bullet_0 \gamma \}$$

$$\operatorname{Der}(t) = \{(\alpha, \beta, \gamma) : t \bullet_2 \alpha + t \bullet_1 \beta = t \bullet_0 \gamma \}$$

left nucleus

mid nucleus (adjoint)

right nucleus

centroid

derivation

t \bullet_2 \alpha: U_2 \times U_1 \rightarrowtail U_0 \\ (u,v) \mapsto t(\alpha(u), v)

t \bullet_1 \beta: U_2 \times U_1 \rightarrowtail U_0 \\ (u,v) \mapsto t(u, \beta(v))

t \bullet_0 \gamma: U_2 \times U_1 \rightarrowtail U_0 \\ (u,v) \mapsto \gamma(t(u, v))

$ \mathcal{LMR}(t) $

Results for nuclei

Definition ($ \boxtimes $)

Let $\alpha_{\ast} = (\alpha_2, \alpha_1, \alpha_0) $, and $ f_{\ast} = (f_2, f_1, f_0) $, where $ \alpha_i \in \operatorname{End}(U_i) $, and $f_i \in \operatorname{End}(V_i) $

Corollary (Wilson Propositions 7.8 and 7.9, "Decomposing $p$-groups via Jordan Algebras")

Let $d: U \times U \rightarrowtail C$ be a nondegenerate Hermitian $C$-form and $b: V \times V \rightarrowtail W $ a $k$-bilinear map. Then $ \operatorname{Adj}(d \otimes b) = \operatorname{Adj}(d) \otimes \operatorname{Adj}(b) $.

Define $ \alpha_{\ast} \boxtimes f_{\ast} \coloneqq (\alpha_2 \otimes f_2, \alpha_1 \otimes f_1, \alpha_0 \otimes f_0) $, where $\alpha_i \otimes f_i \in \operatorname{End}(U_i \otimes V_i) $

Theorem C (L.)

Let $s: U_2 \times U_1 \rightarrowtail U_0$ and $t: V_2 \times V_1 \rightarrowtail V_0 $ be fully non-degenerate bilinear maps. Then

$$\mathcal{LMR}(s \otimes t) = \mathcal{LMR}(s) \boxtimes \mathcal{LMR}(t)$$

Recall $\mathcal{LMR}(s) = \mathcal{L}(s) \oplus \mathcal{M}(s) \oplus \mathcal{R}(s) $

(s \otimes t) \bullet_2 x = \underbrace{ \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_2\!\alpha_1 \!\! & \!\! s\!\bullet_2\!\alpha_2 \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] }_{\displaystyle A} \underbrace{ \left[ \begin{array}{ccc} \phantom{0} & t \bullet_2 f_1 & \phantom{0} \\ \hline \phantom{0} & t \bullet_2 f_2 & \phantom{0} \end{array} \right] }_{\displaystyle X} = \underbrace{ \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_1\!\beta_1 \!\! & \!\! s\!\bullet_1\!\beta_2 \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] }_{\displaystyle B} \underbrace{ \left[ \begin{array}{ccc} \phantom{0} & t \bullet_1 g_1 & \phantom{0} \\ \hline \phantom{0} & t \bullet_1 g_2 & \phantom{0} \end{array} \right] }_{\displaystyle Y} = (s \otimes t) \bullet_1 y

Suppose $ (x,y) \in \mathcal{M}(s \otimes t) $, meaning $ (s \otimes t) \bullet_2 x = (s \otimes t) \bullet_1 y $

Decompose $ x = \alpha_1 \otimes f_1 + \alpha_2 \otimes f_2 $ and $ y = \beta_1 \otimes g_1 + \beta_2 \otimes g_2 $

$ \cong $

$ \otimes $

Recall $x \in \operatorname{End}(U_2 \otimes V_2) \cong \operatorname{End}(U_2) \otimes \operatorname{End}(V_2) $

\underbrace{ \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_2\!\alpha_1 \!\! & \!\! s\!\bullet_2\!\alpha_2 \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] }_{\displaystyle A} M = \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_2\!\tilde{\alpha_1} \!\! & \!\! s\!\bullet_2\!\tilde{\alpha_2} \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] = \underbrace{ \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_1\!\beta_1 \!\! & \!\! s\!\bullet_1\!\beta_2 \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] }_{\displaystyle B}

$ (\tilde{\alpha}_i, \beta_i) \in \mathcal{M}(s) $

M^{-1} \underbrace{ \left[ \begin{array}{ccc} \phantom{0} & t \bullet_2 f_1 & \phantom{0} \\ \hline \phantom{0} & t \bullet_2 f_2 & \phantom{0} \end{array} \right] }_{\displaystyle X} = \left[ \begin{array}{ccc} \phantom{0} & t \bullet_2 \tilde{f_1} & \phantom{0} \\ \hline \rule{0pt}{1.1em} \phantom{0} & t \bullet_2 \tilde{f_2} & \phantom{0} \end{array} \right] = \underbrace{ \left[ \begin{array}{ccc} \phantom{0} & t \bullet_1 g_1 & \phantom{0} \\ \hline \phantom{0} & t \bullet_1 g_2 & \phantom{0} \end{array} \right] }_{\displaystyle Y}

$ (\tilde{f}_i, g_i) \in \mathcal{M}(t) $

\underbrace{ \begin{bmatrix} 3&1\\4&2 \end{bmatrix} }_{\displaystyle M^{-1}} \underbrace{ \left[ \begin{array}{cc} 1 & 2 & 0\\ \hline 0 & 1 & 1 \end{array} \right] }_{\displaystyle X} = \underbrace{ \begin{bmatrix} 3 & 2 & 1\\ \hline 4 & 0 & 2 \end{bmatrix} }_{\displaystyle Y}

$\exists M$ where $AM = B$ and $M^{-1}X = Y$

\underbrace{ \left[ \begin{array}{c|c} 1 & 0 \\ 2 & 1 \\ 0 & 1 \end{array} \right] }_{\displaystyle A} \underbrace{ \begin{bmatrix} 1&2\\3&4 \end{bmatrix} }_{\displaystyle M} = \underbrace{ \left[ \begin{array}{c|c} 1 & 2\\ 0 & 3\\ 3 & 4 \end{array} \right] }_{\displaystyle B}

$AX = BY $ minimal rank factorizations

A = \left[ \begin{array}{c|c} 1 & 0 \\ 2 & 1 \\ 0 & 1 \end{array} \right]

X = \left[ \begin{array}{ccc} 1 & 2 & 0 \\ \hline 0 & 1 & 1 \end{array} \right]

B = \left[ \begin{array}{c|c} 1 & 2 \\ 0 & 3 \\ 3 & 4 \end{array} \right]

Y = \left[ \begin{array}{ccc} 3 & 2 & 1 \\ \hline 4 & 0 & 2 \end{array} \right]

Upshot: global equality leads to local equalities

Field is $ \mathbb{F}_5 $

$ \operatorname{col}(A)=\operatorname{col}(AX)=\operatorname{col}(BY)=\operatorname{col}(B) $

If all nonzero, then up to scalars, either

$$a=b=c \text{ and } x+y=z$$

$$a+b=c \text{ and } x=y=z$$

(Grad student's dream)

$ \operatorname{Der}(s \otimes t) = \operatorname{Der}(s) \boxtimes \operatorname{Der}(t) $

But alas, not the case.

Consider $a \otimes x + b \otimes y = c \otimes z$, a triple in $ \operatorname{Der}(s \otimes t) $

In $ \operatorname{Der}(s) \boxtimes \operatorname{Der}(t) $ means that both

$ (a,b,c) \in \operatorname{Der}(s) $, so $ (a+b=c) $
$ (x,y,z) \in \operatorname{Der}(t)$, so $ (x+y=z) $

Why?

Must have shared factor for sum of two pure tensors to be rank 1

Instead, term in either $ \mathcal{C}(s) \boxtimes \operatorname{Der}(t) $ or $ \operatorname{Der}(s) \boxtimes \mathcal{C}(t) $

The nuclei case applies when one term is zero

Not possible as $ c \otimes z = (a+b) \otimes (x + y) $ has cross terms

Theorem D (L.)

Let $s: U_2 \times U_1 \rightarrowtail U_0$ and $t: V_2 \times V_1 \rightarrowtail V_0 $ be fully non-degenerate bilinear maps.

Then

\operatorname{Der}(s \otimes t) = \operatorname{Der}(s) \boxtimes \mathcal{C}(t) + \mathcal{C}(s) \boxtimes \operatorname{Der}(t) + \iota(\mathcal{LMR}(s) \boxtimes \mathcal{LMR}(t))

Results for derivation

Corollary (Benkart-Osborn, Corollary 4.9, Derivations and Automorphisms of Nonassociative Matrix Algebras)

Let $A$ be a unital algebra. Then $ \operatorname{Der}(\mathbb{M}_n(A)) = I_n \otimes \operatorname{Der}(A) + \operatorname{ad}(\mathbb{M}_n(\operatorname{Nuc}(A))) $

Recall for $u \in \operatorname{End}(A)$, the adjoint action $\operatorname{ad}(u) \in \operatorname{Der}(A) $ is $ x \mapsto ux - xu $

A Locally Independent Unified (LIU)-decomposition of $ (p,q,r) $ consists of a natural number $n$ and vectors $a_i \in A, x_i \in X, b_i \in B, y_i \in Y, c_i \in C, z_i \in Z$ such that

(Decomposition Equality) $ p = \sum_{i=1}^{n} a_i \otimes x_i$, $q = \sum_{i=1}^{n} b_i \otimes y_i $, and $r = \sum_{i=1}^{n} c_i \otimes z_i $
(Local Equality) $\forall i \in [n]$, $a_i \otimes x_i + b_i \otimes y_i = c_i \otimes z_i $

Key definition

p=\color{blue}{a_1\otimes x_1}+\color{green}{a_2\otimes x_2}, \qquad \color{black}{q}=\color{blue}{b_1\otimes y_1}+\color{green}{b_2\otimes y_2}, \qquad \color{black}{r}=\color{blue}{c_1\otimes z_1}+\color{green}{c_2\otimes z_2}.

Example

$p+q=r$, with

\color{blue}{a_1\otimes x_1+b_1\otimes y_1=c_1\otimes z_1}, \qquad \color{green}{a_2\otimes x_2+b_2\otimes y_2=c_2\otimes z_2}.

Local equalities:

p = \underbrace{ \left[\!\!\!\! \begin{array}{c|c} \phantom{0} & \phantom{0} \\ s\!\bullet_2\!\alpha_1 \!\! & \!\! s\!\bullet_2\!\alpha_2 \\ \phantom{0} & \phantom{0} \end{array} \!\!\!\!\right] }_{\displaystyle A} \underbrace{ \left[ \begin{array}{ccc} \phantom{0} & t \bullet_2 f_1 & \phantom{0} \\ \hline \phantom{0} & t \bullet_2 f_2 & \phantom{0} \end{array} \right] }_{\displaystyle X}

Definition

Let $p \in A \otimes X \leq U \otimes V$, $q \in B \otimes Y \leq U \otimes V$, and $r \in C \otimes Z \leq U \otimes V$ satisfy $p+q=r$.

Lemma: Let $ (p,q,r) $ be a triple satisfying $p+q=r$. Then $ (p,q,r) $ has a LIU-decomposition.

Idea: Decompose $U = \bigoplus_{i=1}^{6} U^{(i)} $ and $V = \bigoplus_{i=1}^{6} V^{(i)} $ (e.g $U^{(1)} = A \cap B \cap C $)

in aligned basis

source: https://gist.github.com/cliu587/0db323a55bec8f2d014e870844a18d71

$p+q=r$ over $ \mathbb{F}_{997} $

Fix to Grad Student's Dream

Theorem: There exists examples for $p \in A \otimes X \leq U \otimes V$, $q \in B \otimes Y \leq U \otimes V$, $r \in C \otimes Z \leq U \otimes V$, and $s \in D \otimes W \leq U \otimes V$ satisfying $p+q+r$ without a LIU-decomposition.

Proof:

Work over $ \mathbb{F} = \mathbb{F}_5 $, and let $U,V = \mathbb{F}^4 $, with standard bases

$ D = \text{span}\{e_1,e_2\} \leq U, W = \text{span}\{f_1,f_2\} \leq V $

$ A = \text{span}\{e_1 + e_3, e_2 + e_4 \}, B =\text{span}\{e_1 - e_3, e_2 - e_4 \}, C = \text{span}\{e_1 + e_4, e_2 + 4e_3+ 4e_4 \} $

$ X = \text{span}\{f_1 + f_3, f_2 + f_4 \}, Y =\text{span}\{f_1 - f_3, f_2 - f_4 \}, Z = \text{span}\{f_1 + 4f_4, f_2 + f_3+ 4f_4 \} $

As notation, we shall denote $A = \text{span}\{e_1+e_3 \eqqcolon a_1, e_2+e_4 \eqqcolon a_2\}, B = \text{span}\{b_1,b_2\}$, and so on

Then let

$ p = a_1 \otimes x_1 + a_1 \otimes x_2 + 4 a_2 \otimes x_1 + 2 a_2 \otimes x_2 $
$ q = 3 b_1 \otimes y_1 + 2 b_1 \otimes y_2 + 3 b_2 \otimes y_1$
$ r = 2 c_1 \otimes z_1 + 2 c_1 \otimes z_2 + 3c_2 \otimes z_1 + 4 c_2 \otimes z_2 $
$ s = e_1 \otimes f_1 + e_2 \otimes f_2 $

This quadruple satisfies $p+q+r = s$, with $p \in A \otimes X, q \in B \otimes Y, r \in C \otimes Z, s \in D \otimes W $, and no local equality exists because any nonzero element of $(A \otimes X + B \otimes Y + C \otimes Z) \cap (D \otimes W)$ is rank 2. (Magma calculation)

This example was created so elements in the intersection has to satisfy

\begin{bmatrix} P+Q+R = S & P-Q+RJ = 0 \\ P-Q+JR = 0 & P+Q+JRJ = 0 \end{bmatrix}

And choosing $J$ such that $S \in \mathbb{F}[J] $ is invertible

SKIP DUE TO TIME

If $(x,y) \in \mathcal{M}(s \otimes t) $, the equation $ (s \otimes t)\bullet_2 x = (s \otimes t) \bullet_1 y $ is matrix equality by $ (*) $

For $AX = (s \otimes t) \bullet_2 x = (s \otimes t) \bullet_1 y = BY$ minimal rank factorizations, each column of $A$ is in $\operatorname{CS}(B) $, and each row of $X$ is in $\operatorname{RS}(Y)$

Key idea for nuclei proof

Extras that I cut

Corollary (Brešar, Theorem 3.1, Derivations of Tensor Products of Nonassociative Algebras)

Let $R$ and $S$ be non-associative algebras.

Then every derivation of $R \otimes S$ can be written as $d = \operatorname{ad}(u) + \sum_{j=1}^{p} \lambda_{z_j} \otimes f_j + \sum_{i=1}^{q} g_i \otimes \lambda_{w_i} $, where $u \in \operatorname{Nuc}(R) \otimes \operatorname{Nuc}(S) $, $z_j \in Z(R)$, $f_j \in \operatorname{Der}(S) $, and $g_i \in \operatorname{Der}(R) $

Corollary - $ \mathcal{C}(s \otimes t) = \mathcal{C}(s) \boxtimes \mathcal{C}(t) $

\operatorname{Der}(s \otimes t) = \operatorname{Der}(s) \boxtimes \mathcal{C}(t) + \mathcal{C}(s) \boxtimes \operatorname{Der}(t) + \\ \qquad \qquad \qquad \qquad \iota_{20}(\mathcal{L}(s) \boxtimes \mathcal{L}(t)) + \iota_{10}(\mathcal{R}(s) \boxtimes \mathcal{R}(t)) + \iota_{21}(\mathcal{M}(s) \boxtimes \mathcal{M}(t)) \\[0.5em] \qquad \qquad \qquad \qquad \scriptsize{\text{where } \iota_{20}(\alpha, \gamma) = (\alpha, 0, \gamma), \iota_{10}(\beta,\gamma) = (0,\beta,\gamma), \iota_{21}(\alpha, \beta) = (-\alpha, \beta, 0)}

In summary

Tensors are multilinear maps whose algebraic invariants discern structure
Theorem A, Theorem B - faster algorithm for adjoint and derivation
Theorem C, Theorem D - decomposition theorems on the nuclei and derivation

$ I $

$ J $

$ M_R $

$ M_S $

Algorithms for adjoint and derivation algebra of higher valence tensors (e.g trilinear maps)
Weaken regular family and black-box restriction requirements using randomized computation model
Decomposing the product of bilinear maps

Future directions

Questions?

Tensors are multilinear maps whose algebraic invariants discern structure
Theorem A, Theorem B - faster algorithm for adjoint and derivation
Theorem C, Theorem D - decomposition theorems on the nuclei and derivation