2021 James B. Wilson, Colorado State University
Uriya First
U. Haifa
Joshua Maglione,
Bielefeld
Peter Brooksbank
Bucknell
(I hope!)
Are all of these tensors?
If you don't use linear combinations on some axis of your data...
then its not actually a tensor, sorry.
"What is a vector?"
"An element of a vector space."
\(U_0\oslash U_1=\{f:U_1\to U_0\mid f(u+\lambda v)=f(u)+\lambda f(v)\}\)
\(U_0,U_1,\ldots\) are vector spaces (or modules).
Linear maps:
\(U_0\oslash U_1\oslash U_2 := \{f:U_2\to U_0\oslash U_1 \mid f(u+\lambda v)=f(u)+\lambda f(v)\}\)
Bi-Linear maps:
\(U_0\oslash \cdots U_{k-1}\oslash U_k:=(U_0\oslash\cdots\oslash U_{k-1})\oslash U_k\)
\(k\)-multi-linear maps:
"What is a tensor?"
"An element of a tensor space."
\(U_0,U_1,\ldots\) are vector spaces (or modules).
\(U_0\oslash \cdots U_{k-1}\oslash U_k:=(U_0\oslash\cdots\oslash U_{k-1})\oslash U_k\)
\(k\)-multi-linear maps:
Defn. A Tensor Space \(T\) is a vector space an a linear map \[\langle \cdot|:T\to U_0\oslash\cdots\oslash U_k\]
Tensors are elements of tensor spaces.
\(T=\mathbb{M}_{2\times 3}(\mathbb{R})\) is a tensor space in at least 3 ways!
\[\langle \cdot |:T\mapsto \mathbb{R}^2\oslash \mathbb{R}^3\]
\[\langle M|u\rangle := Mu\]
\[|\cdot \rangle:T\mapsto \mathbb{R}^3\oslash \mathbb{R}^2\]
\[\langle v| M\rangle := v^{\dagger}M\]
\[|\cdot|:T\mapsto \mathbb{R}\oslash\mathbb{R}^2\oslash \mathbb{R}^3\]
\[\langle v| M|u\rangle := v^{\dagger}Mu\]
Matrix as linear map on right.
Matrix as linear map on left.
Matrix as bilinear form.
This abstraction does wonders for creating a fluid tensor software package.
low-brow: reindex,
high-brow: affine transforms of polytopes
Evaluation
Contractions
Layout data so nothing moves!
Logically equivalent circuits
Fight eager evaluation
=
(Data) Acting on Tensors as arrays
(Lin. Alg.) Acting on tensors as functions
(Physics/Algebra) Acting on Tensors as Operads/Networks
Generalizes Characteristic polynomial to ideals
\[I(t,\omega)=(x^2-x, y^2-y, xy)\qquad I(t,\tau)=(x^2, y^3,xy-y^2)\]
Multi-spectrum Rule = \(\langle t| p(\omega) =0\)
Thm FMW-Connection. \(S\) set of tensors, \(P\subset \mathbb{R}[X]\), \(\Omega\subset \prod_a \mathbb{M}_{d_a}(\mathbb{R})\)
\[T(P,\Omega)=\{t\mid P\textnormal{ in multi-spec } t\text{ at } \Omega\}\]
\[I(S,\Omega)=\{p\mid p\textnormal{ in multi-spec } S\text{ at } \Omega\}\]
\[Z(S,P)=\{\omega \mid P\textnormal{ in multi-spec } S\text{ at } \omega\}\]
Then \[S\subset T(P,\Omega)\Leftrightarrow P\subset I(S,\Omega) \Leftrightarrow \Omega\subset Z(S,P)\]
Thm FMW-Construction.
These are each polynomial time computable.
Multi-spectrum Rule = \(\langle t| p(\omega) =0\)
Functors on tensors, e.g. \((U_0\oslash\cdots \oslash U_K)\to (U_i\otimes\cdots\otimes U_k\to U_0\oslash\cdots\oslash U_{i-1})\)
Save yourself time if you program these functors and avoid boiler plate later
Red is the space we search/work within.
Add some algebra \(A\) in the form of \(U\otimes_A V\)
The bigger the algebra the better.
Rule of thumb
\[\dim (U\otimes_A V)\approx \frac{\dim U\dim V}{\dim A}\]
Some effort now in working with the algebra & modules, yet you can at least prove and plan for that.
Theorem Brooksbank-W. (2012) \[\mathrm{Adj}(S)=\{(F,G)\in \mathbb{M}_a(\mathbb{R})\times \mathbb{M}_b(\mathbb{R})\mid (\forall T\in S)(FT=TG^t)\}\] is an optimal choice and unique up to isomorphism.
\(\mathbb{R}^4\otimes_{\mathbb{R}}\mathbb{R}^{12}\otimes_{\mathbb{R}}\mathbb{R}^{6}\)
\(\mathbb{R}^4\otimes_{\mathbb{M}_2(\mathbb{R})}\mathbb{R}^{12}\otimes_{\mathbb{M}_3(\mathbb{R})}\mathbb{R}^{6}\)
\(\cong\mathbb{R}^2\otimes_{\mathbb{R}}\mathbb{R}^{2}\otimes_{\mathbb{R}}\mathbb{R}^{2}\)
\(\mathbb{R}^4\otimes_{\mathbb{M}_4(\mathbb{R})}\mathbb{R}^{12}\otimes_{\mathbb{M}_3(\mathbb{R})}\mathbb{R}^{6}\)
\(\cong\mathbb{R}\otimes_{\mathbb{R}}\mathbb{R}\otimes_{\mathbb{R}}\mathbb{R}^{2}\)
\(\mathbb{R}^4\otimes_{\mathbb{M}_2(\mathbb{R})}\mathbb{R}^{12}\otimes_{\mathbb{M}_6(\mathbb{R})}\mathbb{R}^{6}\)
\(\cong\mathbb{R}^2\otimes_{\mathbb{R}}\mathbb{R}\otimes_{\mathbb{R}}\mathbb{R}\)
Why can't we just act on one side?
E.g. \(U\otimes_A V\) needs \(U_A, {_A V}\). Worse, \(U\otimes_A V\otimes_B W\) needs \({_A V_B}\) a "bi-module".
Why do we tolerate "natural" isomorphisms \[U\otimes (V\otimes W)\cong (U\otimes V)\otimes W\]
If its natural, can't we just write these down as equal?!
Whitney Tensor Product
A Different Tensor Product
New Tensor product:
\(\Omega\subset \mathbb{M}_{d_1}(\mathbb{R})\times\cdots \times \mathbb{M}_{d_k}(\mathbb{R})\); \(P\subset \mathbb{R}[x_1,\ldots,x_k]\)
\[\Xi(P,\Omega)=\left\langle \sum_e \lambda_e \omega_1^{e_1}\otimes\cdots\otimes \omega_k^{e_k} ~\middle|~\sum_e\lambda_e X^e\in P, \omega\in \Omega\right\rangle\]
\[(]U_1,\ldots,U_k[)_{\Omega}^P := (U_1\otimes \cdots \otimes U_k)/\Xi(P,\omega)\]
Then we have:
\[(]\cdots[):U_1\times\cdots\times U_k\hookrightarrow (]U_1,\ldots,U_k[)_{\Omega}^P\]
defined by
\[(]u_1,\ldots,u_k[) := u_1\otimes\cdots\otimes u_k+\Xi(P,\Omega)\]
Condensing Whitney Tensor Products
Condensing our alternative
One corner to contract makes each axis independent.
(No bimodules, no "associative" rules)
\[(]\cdots[):U_1\times\cdots\times U_k\hookrightarrow (]U_1,\ldots,U_k[)_{\Omega}^P\]
is the universal tensor such that every \(\omega\in \Omega\) has \(P\) in its multi-spectrum.
Intuition....force the spectrum
Consequence:
Theorem First-Maglione-W. \(\mathrm{Der}(t)\) is all \((\delta_i)_i\in \prod\mathbb{M}_{d_i}(\mathbb{R})\) satisfying \[0=\langle t|\delta_1u_1,\ldots,u_n\rangle+\cdots+\langle t|u_1,\ldots,\delta_v u_v\rangle\] is an optimal choice and unique up to isomorphism.
Theorem First-Maglione-W. If \(P=(\Lambda X)\), \(\Lambda\in \mathbb{M}_{r\times k}\) is full rank and if \[Z(t,P)=\{\omega\mid P\textnormal{ in the multi-spec } t\}\]
is an algebra, then it is a Lie algebra in at least \[k-2r\] coordinates.
\(U\otimes_A V\) for \(A\) associative is a fluke, it is the r=1 case when k=2.
Problem posed in:
Acar, Camtepe, and Yener, Collective Sampling and Analysis of High Order Tensors for Chatroom Communications, Proc. 4th IEEE Int. Conf.Intel. and Sec. Info., 2006, pp. 213–224
Reality
The algebra is never there,
never that nice,
not even associative.
No algebra? Make one by enrichment!
Its decompositions do the job.
\(t\in \mathbb{R}^{10}\otimes \mathbb{R}^7\)
\(\mathrm{Adj}(t)\cong \mathbb{R}\oplus \mathbb{M}_2(\mathbb{R})\) \[\begin{aligned} t & \in \mathbb{R}^{10}\otimes_{\mathrm{Adj}(t)}\mathbb{R}^7 \\ & \cong (\mathbb{R}^{10}\otimes_{\mathbb{R}\oplus 0}\mathbb{R}^7)\oplus (\mathbb{R}^{10}\otimes_{0\oplus \mathbb{M}_2(\mathbb{R})}\mathbb{R}^7) \\ & = (\mathbb{R}^2\otimes \mathbb{R}^3)\oplus (\mathbb{R}^4\otimes\mathbb{R}^2)\end{aligned}\]
\[U\otimes_{A_1\oplus A_2}V\otimes W\cong (U_1\otimes_{A_1} V_1\otimes W)\oplus(U_2\oplus_{A_2}V_2\otimes W)\]
Thm. (FMW-Singular)
Valence 2
Valence 3
Parker-Norton 1975 MeatAxe: polynomial time algorithm for \[XTX^{-1}=T_1\oplus \cdots \oplus T_{\ell}.\]
Performance: Dense 1/2 million dimensions in an hour, on desktop.
W. 2008: Proved uniqueness and polytime-algorithms for \[\begin{aligned} XTX^{\dagger} & = T_1\perp\cdots\perp T_{\ell}\\ XTY & =T_1\oplus \cdots\oplus T_{\ell}\end{aligned}\]
Generalizations being explored now.
Pros.
Cons.
Data credit to Frank W. Marrs III
Los Alamos National Labs
Further Fact of spectra:
If derivation and nilpotent then...
How to apply?
Partners
Action/Reaction
Benafactor
Between pairs, 6 total (3 pictured)
\(\vdots\)
Tensor:
Tensor:
Verstraete, Dehaene, De Moor, Verschelde, Four qubits can be entangled in nine different ways, Phys. Rev. A 65 (2002)
D. and B. Williamson,Mari¨en, Matrix product operators for symmetry-protected topological phases: Gauging and edge theories, Phys. Rev. B 94 (2016)
Quantum Particles modeled as vectors in \(\mathbb{C}^d\)
Entangled Particles as in \(\mathbb{C}^{d_1}\otimes\cdots \otimes \mathbb{C}^{d_k}\)
Visualize as n-gon.
Objective: What is the large-scale physics of a many body quantum material?
Comes down to symmetries of then tensors.
Valence 4?
What qualifies as a symmetry of a tensor? Not just anything...surprisingly combinatorial...
Valance 3
Yes
No.
Thm FMW-Groupoid.
\[Z(t,p)^{\times}=\{\omega\mid p \textnormal{ in multi-spec } t\}\] is a group in some tensor category if, and only if, \[p=X^g(X^e-X^f)\] where \(e,f\) have disjoint support and are \(\{0,1\}\) valued.
Solution: chase the algebraic geometry of the spectra.... it turns out to be toric and thus combinatorial!
Solving \((\forall i)(XA_i+B_iY=C_i)\) in nearly linear time
\((\forall i)(XA_i+B_iY=C_i)\) and variations.
Solving \((\forall i)(XA_i+B_iY=C_i)\) is linear in \(d^2\) variables so \(O(d^{2\omega})\subset O(d^6)\) work.
Good enough in theory, but hard to fit in memory and unrealistic at scale.
Yields \(O(d^{\omega})\) time algorithms, \(\omega\leq 3\)
\[\delta_A^{12}(u\otimes v\otimes w)=u\otimes v\otimes w-u_1\sum_{\ell=2} e_{\ell}\otimes e_{\ell}A v\otimes w\]
Prop. \(\delta_A^{12}\circ \delta_B^{13}=\delta_B^{13}\circ \delta_A^{12}\)
\(E=\delta_B^{13}\) and \(F=\delta_A^{12}\)
Face Elimination: a tensor solution
Thm Collery-Maglione-W.
QuickSylver solves simultaneous generalized Sylvester equations in time \(O(d^{3})\) (for 3-tensors).
Several related videos/software/resources at
https://thetensor.space/
A recently updated version of some of the main results at
https://www.math.colostate.edu/~jwilson/papers/Densor-Final-arxiv.pdf