Paper I visual guide · Visual 5 of 22 · PRESENTATIONS / WHY A LEFT IDEAL APPEARS

Relations identify words, not just outputs

Main text: Theorem 3.2, Corollary 3.4, and Proposition 3.5

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Let \(F={\Bbbk}\langle z_1,\ldots,z_r\rangle\) be the algebra of noncommuting operator words. Evaluation at a fixed seed is

\[ \Phi_s:F\to X,\qquad f\mapsto f(T_1,\ldots,T_r)s. \]

Two words represent the same generated vector exactly when their difference lies in \(I_s={\operatorname{Ker}}\Phi_s\). Consequently

\[ F/I_s\cong{\operatorname{Hull}}_F(s) \]

as left modules. The relation \(f(T)s=0\) implies \(g(T)f(T)s=0\) for every word \(g\), which explains closure under multiplication on the left.

The distinction is visible in two dimensions

Use \(X={\mathbb Q}^2\), seed \(e_1=(1,0)^\mathsf T\), and the full matrix algebra. Its seed annihilator is the set of matrices with zero first column:

\[ I_{e_1}=\left\{{\begin{pmatrix}0&a\\0&b\end{pmatrix}}:a,b\in{\mathbb Q}\right\}. \]
Right multiplication can destroy a seed relation
Original drawing. Read the description below; expand the labels and data for selectable text.

Reading the drawing

Three matrices display C, B and their product in that order. The source labels show that C kills the seed while CB returns the seed, illustrating the difference between a left ideal and a two-sided ideal.

Drawing labels and data 19 labels

Read the original labels as selectable text. Tables preserve grid relationships; other short labels are grouped compactly.

\(C=E_{12}\)
RowColumn 1Column 2
Row 1\(0\)\(1\)
Row 2\(0\)\(0\)
\(B=E_{21}\)
RowColumn 1Column 2
Row 1\(0\)\(0\)
Row 2\(1\)\(0\)
\(CB=E_{11}\)
RowColumn 1Column 2
Row 1\(1\)\(0\)
Row 2\(0\)\(0\)
  • \(\times\)
  • \(=\)
  • \(Ce_1=0\)
  • \(CBe_1=e_1\)

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Evidence note from the source edition

The relation \(Ce_1=0\) is destroyed by multiplying \(C\) on the right by \(B\). The product \(CB\) sends the seed to itself.

For arbitrary \(D\), \((DC)e_1=D(Ce_1)=0\). But \((CB)e_1=e_1\ne0\). Thus \(I_{e_1}\) is a left ideal, but it is not stable under multiplication on the right. The quotient retains the action on generated vectors. Calling it a ring would lose the distinction that makes the presentation correct.

Several seeds give a module of relations

For \((s_1,\ldots,s_m)\), evaluation is \((f_1,\ldots,f_m)\mapsto\sum_j f_j(T)s_j\). Its kernel is a submodule of \(F^m\) and can include relations between seeds, not only relations annihilating each seed separately.

Read this as

A presentation says which operator expressions are identified in a particular realization. A seed annihilator is not automatically an algebra relation ideal.

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