Equivalent odd-subspace formulation of the Chowla–Milnor conjecture

From papers

Let k2k\geqslant 2 and q3q\geqslant 3 be integers. For 1a<q1\leqslant a<q with gcd(a,q)=1\gcd(a,q)=1, define

ζ+(k,aq):=ζ(k,aq)+(1)kζ(k,1aq),\zeta^{+}\left(k,\frac{a}{q}\right):=\zeta\left(k,\frac{a}{q}\right)+(-1)^k\zeta\left(k,1-\frac{a}{q}\right), ζ(k,aq):=ζ(k,aq)(1)kζ(k,1aq).\zeta^{-}\left(k,\frac{a}{q}\right):=\zeta\left(k,\frac{a}{q}\right)-(-1)^k\zeta\left(k,1-\frac{a}{q}\right).

Define

Vk(q):=SpanQ{ζ(k,aq)1a<q2, gcd(a,q)=1}V_k^{-}(q):=\operatorname{Span}_{\mathbb{Q}}\left\{\zeta^{-}\left(k,\frac{a}{q}\right)\mid 1\leqslant a<\frac q2,\ \gcd(a,q)=1\right\}

and

Vk+(q):=SpanQ{ζ+(k,aq)1a<q2, gcd(a,q)=1}.V_k^{+}(q):=\operatorname{Span}_{\mathbb{Q}}\left\{\zeta^{+}\left(k,\frac{a}{q}\right)\mid 1\leqslant a<\frac q2,\ \gcd(a,q)=1\right\}.

Equivalent Chowla–Milnor formulation. One has

dimQVk(q)=φ(q)2\dim_{\mathbb{Q}}V_k^{-}(q)=\frac{\varphi(q)}2

and

Vk+(q)Vk(q)={0}.V_k^{+}(q)\cap V_k^{-}(q)=\{0\}.

Since the even subspace is known to have dimension φ(q)/2\varphi(q)/2 and Vk(q)=Vk+(q)+Vk(q)V_k(q)=V_k^{+}(q)+V_k^{-}(q), these conditions are equivalent to the Chowla–Milnor linear independence conjecture. The source says that, apart from the case k=2k=2, q=3q=3, other cases remain open.

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Sources & referencesView supporting material

Primary source

Li Lai and Jia Li, “A partial result towards the Chowla–Milnor conjecture”, arXiv:2505.12687 (2026).

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