Direct Sum Decomposition compatible with the equation \( p+q=r \) in \(U \otimes V \)

These slides are a record keeping tool and will not be used at for any actual presentations, hence the tiny font size and color choices at times

Throughout \( U \) and \( V \) will be finite dimensional vector spaces 

Consider the equation \( p+q=r \) in \( U \otimes V \), with \(U\) and \(V\) over \( \text{GF}(997) \)

Consider the equation \( p+q=r \) in \( U \otimes V \)

Goal (compatible decomposition): choose decompositions \(U = \bigoplus_{i=1}^{6} U_i\) and \(V = \bigoplus_{j=1}^{6}V_j\) such that

(Less ambitious) Compatible expansion goal that's still sufficient for derivation of product of tensors result

Find \(a_i, b_i, c_i, x_i,y_i,z_i\) such that

• \( \forall i,\; a_i \in A, b_i \in B, c_i \in C, x_i \in X, y_i \in Y, z_i \in Z \)
• \( \forall i,\; a_i \otimes x_i + b_i \otimes y_i = c_i \otimes z_i \)
• \( \sum_i a_i \otimes x_i = p, \sum_i b_i \otimes y_i = q, \sum_k c_i \otimes z_i = r \)

(By decomposing the spaces, we separate out the individual equations into a few different types, categorized by what kind of equations can satisfy \( a \otimes x + b \otimes y = c \otimes z\))

Compatible decomposition implies compatible expansion as the \( (U5\oplus U6) \otimes (V5 \oplus V6) \) bottom right corner is empty so we can expand on the left/right basis strategically

\( p \in A \otimes X \)

Let subspaces \(A,B,C \leq U\), and subspaces \(X,Y,Z \leq V \) be of minimum dimension such that 

\( q \in B \otimes Y \)

\( r \in C \otimes Z \)

Recall \( p,q,r \in U \otimes V \) such that \( p+q=r \)

• \( U_1 = A \cap B \cap C \)
• \( U_2 = \text{complement of } U_1 \text{ in } A \cap B \)
• \( U_3 = \text{complement of } U_1 \text{ in } A \cap C \)
• \( U_4 = \text{complement of } U_1 \text{ in } B \cap C \)
• \( U_5 = \text{complement of } A \cap B + A \cap C \text{ in } A \)
• \( U_6 = \text{complement of } A \cap B + B \cap C \text{ in } B \)

First, prove the sum \( \mathcal{U} := U_1 + U_2 + U_3 + U_4\) is direct

Claim: \(A = U_1 \oplus U_2 \oplus U_3 \oplus U_5 \)

Visually

\( (\pi_{U_4 \oplus U_6} \otimes I)(p) = 0 \) and \( (\pi_{U_3 \oplus U_5} \otimes I)(q) = 0 \)

\( (I \otimes \pi_{V_4 \oplus V_6})(p) = 0 \) and \( (I \otimes \pi_{V_3 \oplus V_5})(q) = 0 \)

looks like

Because \( (\pi_{U_3} \otimes I)(q) = 0\), let \(s \coloneqq (\pi_{U_3} \otimes I)(p) = (\pi_{U_3} \otimes I)(r) \)

Then \(s \in U_3 \otimes X \) and \(s \in U_3 \otimes Z \)

Same for \( (\pi_{U_5} \otimes I)(q) = 0\)

Same when \(p\) is zero for a row: \( (\pi_{U_4} \otimes I)(p) = 0\) and \( (\pi_{U_6} \otimes I)(p) = 0\)

Same for projections to columns \(V_3, V_4, V_5, V_6\)

In conclusion

Recall \(p+q=r\) example over \(\text{GF}(997)\)

Takeaway: The equation \(p+q = r \text{ in } U \otimes V \) allows for direct-sum decompositions of \( U \) and \( V \) in a compatible manner

Original motivation

Fix some \( u \in U \) and \( v \in V \)

Let \(x,y,z \in \text{End}(U \otimes V) \) such that

• \(x(u\otimes v) + y(u \otimes v) = z(u \otimes v) \) - a derivation condition
• and \( x = \sum_i \alpha_i \otimes f_i, y = \sum_j \beta_j \otimes g_j, z = \sum_k \gamma_k \otimes h_k \) under identification \( \text{End}(U \otimes V) \cong \text{End}(U) \otimes \text{End}(V) \)

Project to 1-dimensional subspace \(\langle u \rangle \leq U_1\): \( (\pi_{\langle u \rangle} \otimes I)(p) + (\pi_{\langle u \rangle} \otimes I)(q) = (\pi_{\langle u \rangle} \otimes I)(r) \)

Write as \( p_u + q_u = r_u \), and observe \( p_u \in (\langle u \rangle \otimes X) \cong X \), \(q_u \in (\langle u \rangle \otimes Y) \cong Y \), and \( r_u \in (\langle u \rangle \otimes Z) \cong Z \)

This is an equation in \(V\), and symmetrically we get equations in \(U\) projecting to \( \langle v \rangle \leq V_1\)

Similar, project to one dimensional subspaces

(e.g \( \langle u \rangle \leq U_2 \) and \( \langle v \rangle \leq V_2 \), then \( (\pi_{\langle u \rangle} \otimes \pi_{\langle v \rangle})(p) + (\pi_{\langle u \rangle} \otimes \pi_{\langle v \rangle})(q) = 0 \).

Write as \(p_u + q_u = 0 \), and observe \(p_u, q_u \in \langle u \rangle \otimes \langle v \rangle \cong K \) are just scalars

Takeaway: \(p+q =r \text{ in } U \otimes V \) ("2D" equation) is actually a collection of equations on subspaces of \(U\) and \(V\) ("1D" conditions)

(Side goal)

Classify the possible dimensions using \( p+q+r=s \) constraint using Kronecker Canonical Form

\( (I_n, M) \) where M is a full rank linear transformation in JNF  - \(\text{dim}(U_1) = \text{dim}(V_1) = n \)

\( (I_n, J_n(0) ) \) where \( J_n(0) \) is size \(n\) Jordan block for eigenvalue \(0 \)  - \(\text{dim}(U_1) = \text{dim}(V_1) = n-1, \text{dim}(U_3) = \text{dim}(V_3) = 1 \)

\(\epsilon_1\) = \( \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) \)  (size \( 1 \) left minimal index) - \(\text{dim}(V_1) = 1, \text{dim}(U_5) = 1, \text{dim}(U_6) = 1 \)

\(\eta_1\) = \( \left( \begin{bmatrix} 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \end{bmatrix}\right) \)  (size \( 1 \) right minimal index) - \(\text{dim}(U_1) = 1, \text{dim}(V_5) = 1, \text{dim}(V_6) = 1 \)

\(\epsilon_n\) =  (size \( n \) left minimal index, a pair of \((n+1) \times n\) matrices ) - \(\text{dim}(V_1) = n, \text{dim}(U_1) = n-2, \text{dim}(U_2) = \text{dim}(U_3) = \text{dim}(U_4) = 1 \)

\(\eta_n\) =  (size \( n \) right minimal index, a pair of \( n \times (n+1) \) matrices ) - \(\text{dim}(U_1) = n, \text{dim}(V_1) = n-2, \text{dim}(V_2) = \text{dim}(V_3) = \text{dim}(V_4) = 1 \)

All possible dimensions are sums of dimensions of above as Kroncker canonical form is direct sum of these blocks

Key fact: \( C_0 \leq A_0 \oplus B_0 \) gives uniqueness on the \(U_5, U_6\) region

What about \( p+q+r = s \)?

Counterexample to naive shared on one factor, der on the other factor

\(p = a \otimes x \in A \otimes X \)

\(q = b \otimes y \in B \otimes Y \)

\(r = c \otimes z \in C \otimes Z \)

\(s = d \otimes w \in D \otimes W \)

Where

  \( d = e_1 + e_2, w = f_1 + f_2 \),

  \(a = e_1 - e_2, x = f_1 - f_2 \),

  \(b = 2e_1, y = f_2 \),

  \(c = 2e_2, z = f_1 \)

Then \(\{a,b,c,d\} \) span pairwise disjoint spaces, and the same for \( \{x,y,z,w\} \). The relations are \(d = a+c \) and \(w = x + 2y \). (No "centroid" like relation here)

\( (a,c,d) \) are sort of a \(\text{Der}_{310}(s) \) and \( (x,y,w) \) are sort of a \(\text{Der}_{320}(t) \) condition.

This says unlike the 3-tensor case, expanding \( (a_i, b_i, c_i, d_i) \) (or symmetrically \( (x_i,y_i,z_i,w_i) \) ) on appropriate common subspaces will not give a correct expansion. What's needed is a smarter strategy. Try induction? Or frame this technique as fundamentally less visual

TODO:

Lemma 1 (compatible expansion): Given \(p+q+r=s \), there exists \(a_i,b_i,c_i,d_i, x_i,y_i,z_i,w_i \) such that \( \forall i, a_i \otimes x_i + b_i \otimes y_i + c_i \otimes z_i = d_i \otimes w_i \), and \( a_i \in A \) and so forth, and \( \sum_i a_i \otimes x_i = p\) and so forth 

Lemma 2: An equation of the form \( a \otimes x + b \otimes y + c \otimes z = d \otimes w \) can only be of a few forms, all having colinear pieces except one where \(\text{span}\{a,b,c\} \) and \( \text{span}\{x,y,z\} \) are both 2-dimensional. Roughly speaking, each equation is a derivation part combined with a centroid part, except this last case which is combining two non-overlapping derivation parts.

Lemma 3 (compatible decomposition): It is possible to decompose each of the spaces U and V into pairwise intersection cases, 2-dimensional span cases, and the rest of the cases.

Lemma 4: After this splitting the nonzero cells of the compatible decomposition has a structured and predictable pattern that works with the compatible expansion.

Lemma 5: This compatible expansion pattern can be interpreted in context of the derivation algebra of the product of higher valence tensors in a general way. 

For \(p+q=r\) equations, compatible decomposition leads to compatible expansion leads to factor-wise (local) constraints leads to local to global understanding of der. This fact is no longer true for equations of the form \( p+q+r = s \)

Attempts:

• Non-constructive proof of the existence of compatible expansions?
• Non-decomposition based algorithm? (allow for "duplicates")
• Non-algorithmic justifications (allow for "existence" arguments)
• Expand just along 1 direction rather than 2 and use pure tensor argument to relate individual sums
  • Need to expand in a clever fashion to ensure terms are all in the right spaces...

Assuming we can prove the existence of a compatible expansion for equations of the form \( p+q+r=s \), what can we say about translating this expansion into factor-wise constraints?