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

About 21 years old · traced to 2005

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.

References

Primary source

Kalle Karu, “Lefschetz decomposition and the cd-index of fans”, arXiv:math/0509220 (2005).

intersection-cohomology linear-algebra

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Open

A reader-posted argument says the conjecture is trivially true using the zero map, while the injective version needed in the application is false, but neither claim has independent verification.

The conjecture asks whether the lemma remains valid when each self-adjoint map is restricted to multiplication by a constant, meaning $K^\perp\cap L(K)=0$ . No proposer or date is identified.

Posted attempt

The attempt claims a complete proof of the literal conjecture: choosing every constant to be zero gives $L(K)=0$ , hence $K^\perp\cap L(K)=0$ , without using the projection hypotheses. It also claims a concrete counterexample to the stronger invertible or injective interpretation, using $V=(\mathbb{C}^2)^3$ and three scalar blocks; the restricted pairing is asserted to be singular for every choice of constants. These claims have not been independently verified.

Current status (as of August 2026): the literal conjecture is claimed solved by the zero-map observation, while the intended injective strengthening is claimed false; both remain unverified in the public record.

ProofThis solution needs a summarySee full solutionHide full solution

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.