Constant-multiplication conjecture in the self-adjoint linear-map lemma

Let $V=\bigoplus_i V_i$ be a finite-dimensional vector space, with each $V_i$ equipped with a non-degenerate symmetric bilinear pairing, and let $K\subset V$ be a subspace such that $K\subset K^\perp$ . Assume additionally that the projections $V\to V_i$ map $K$ onto $V_i$ for every $i$ . Constant-multiplication conjecture. The conclusion of the preceding lemma remains true when each self-adjoint map $L_i:V_i\to V_i$ is required to be multiplication by some constant $c_i$ . The preceding lemma asserts the existence of self-adjoint maps satisfying $K^\perp\cap L(K)=0$ ; this proposed restriction is used to obtain injectivity in the associated geometric argument, but the source does not establish it.

The conjecture as printed is true trivially, but the injective strengthening needed for its intended application is false.

As stated, there is no requirement that the constants be nonzero or that the resulting map be injective. Choose

c_i=0\qquad\text{for every }i.

Then $L(K)=0$ , and therefore

K^\perp\cap L(K)=0.

This proves the literal conjecture exactly as stated, without even using the additional surjective-projection hypotheses.

However, that conclusion does not provide the injectivity needed by the surrounding geometric argument. In fact, the natural strengthened statement for general invertible scalar-block maps is false.

Work over $\mathbb C$ , let

V=(\mathbb C^2)^3,

and give its three summands the nondegenerate symmetric Gram matrices

J=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad J,\qquad -J.

Define

K= \{((x,z),(y,z),(x+y,z)):x,y,z\in\mathbb C\}.

Every projection from $K$ onto a summand $\mathbb C^2$ is surjective. Writing

\iota(x,y,z)=((x,z),(y,z),(x+y,z)),

one computes

\begin{aligned} \langle\iota(x,y,z),\iota(x',y',z')\rangle ={}&xz'+zx'+yz'+zy'\\ &-(x+y)z'-z(x'+y')=0. \end{aligned}

Hence $K\subseteq K^\perp$ , and since

\dim K=3=\tfrac12\dim V,

in fact $K=K^\perp$ .

For a scalar-block map

L=\operatorname{diag}(c_1I_2,c_2I_2,c_3I_2),

the restricted pairing

(s,t)\longmapsto\langle\iota(s),L\iota(t)\rangle

has Gram matrix

B(c_1,c_2,c_3)= \begin{pmatrix} 0&0&c_1-c_3\\ 0&0&c_2-c_3\\ c_1-c_3&c_2-c_3&0 \end{pmatrix}.

Its determinant vanishes identically. Consequently the induced map

K\longrightarrow V/K^\perp,\qquad k\longmapsto Lk+K^\perp,

is never injective, for any choice of constants.

If all $c_i\neq0$ , then $L$ itself is invertible. Taking a nonzero $t\in\ker B$ yields

0\neq L\iota(t)\in K^\perp\cap L(K).

Explicitly, if

(c_1-c_3,c_2-c_3)\neq(0,0),

one may take

t=(c_2-c_3,-(c_1-c_3),0).

If all three constants are equal and nonzero, then $L(K)=K$ .

Thus the literal conjecture is proved by the zero map, whereas its meaningful generic/invertible interpretation and the injectivity required for the proposed application are disproved by the explicit example above.

Constant-multiplication conjecture in the self-adjoint linear-map lemma

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