The Signed Left Multiplication on the Algebra of Arithmetic Functions

Introduction

Let $\mathcal{A}$ be the algebra of arithmetic functions under Dirichlet convolution, with grade involution $\alpha(f)=\lambda f$, $\lambda$ the Liouville function. For $a\in\mathcal{A}$ the signed left multiplication is the operator $T_a(f)=a*\alpha(f)$; it is the arithmetic form of the map $x\mapsto a\alpha(x)$ of the written theory of a ring with involution. Because $\alpha$ is an automorphism of the commutative algebra, the operator admits two useful readings, $T_a=L_a\alpha=\alpha L_{\alpha(a)}$, and its products are again unsigned left multiplications, $T_aT_b=L_{a*\alpha(b)}$. This article computes the adjoint, the self-adjointness, isometry and unitarity conditions, the kernel and the cokernel, and it records exactly which part of the signed structure survives over a commutative algebra. The signed sandwich of which $T_a$ is the one-sided case is The Signed Sandwich on the Algebra of Arithmetic Functions; the adjoint of the unsigned left multiplication is The Adjoint of the Left Multiplication on the Algebra of Arithmetic Functions, later in this category.

The form of the category is the coefficient form $\langle f,g\rangle=\sum_nf(n)\overline{g(n)}$ on $\mathcal{H}=\ell^2$. The transposed multiplication is $(\Theta_cg)(k)=\sum_{m\ge1}\overline{c(m)}g(mk)$, the adjoint of $L_c$. The algebra $\mathcal{A}$ is an integral domain, being the monoid algebra of the multiplicative monoid $\mathbb{N}$, whose monoid is cancellative and freely generated by the primes. Nothing here reads a distance as an object.

The Operator and its Products

Definition and the two readings

Definition. For $a\in\mathcal{A}$ the signed left multiplication is $$ T_a:\mathcal{A}\to\mathcal{A},\qquad T_a(f)=a*\alpha(f). $$ The unsigned left multiplication is $L_c(f)=c*f$.

Theorem. For every $a$, $$ T_a=L_a\circ\alpha=\alpha\circ L_{\alpha(a)},\qquad \alpha T_a=L_{\alpha(a)},\qquad T_a\alpha=L_a , $$ and the last two identities are the two readings of the operator. The twisted product $$ a\star b:=a*\alpha(b) $$ is associative, $\alpha$ is an isomorphism of algebras $\alpha:(\mathcal{A},\star)\to(\mathcal{A},*)$, and the product of two signed left multiplications is the unsigned left multiplication of the twisted product: $$ T_aT_b=L_{a\star b}=L_{a*\alpha(b)} . $$

Proof. The first display is the definition together with $\alpha L_{\alpha(a)}=L_a\alpha$, which holds because $\alpha(c*f)=\alpha(c)*\alpha(f)$ and $\mathcal{A}$ is commutative. For the product, $T_aT_b=L_a\alpha L_b\alpha=L_aL_{\alpha(b)}\alpha^2=L_{a*\alpha(b)}$. Associativity of $\star$ is $(a\star b)\star c=a*\alpha(b*\alpha c)=a*\alpha(b)*c$, and the same expression is $a\star(b\star c)$; the isomorphism is $\alpha(a\star b)=\alpha(a)*b=\alpha(a)\star\alpha(b)$.

Corollary. $T_a$ is injective if and only if $a\ne0$, and its image is the ideal $\alpha(a)*\mathcal{A}=a*\mathcal{A}$. The family $\{T_a\}$ is closed under products into unsigned left multiplications, and it is commutative exactly when $a*\alpha(b)=b*\alpha(a)$ for all $a,b$, which fails: for $a=\delta_2$, $b=\varepsilon$ one has $a*\alpha(b)=\delta_2$ while $b*\alpha(a)=\lambda\delta_2$.

Proof. $T_a=L_a\alpha$ is the composite of the invertible $\alpha$ with $L_a$, and $L_a$ is injective with image $a*\mathcal{A}$ because $\mathcal{A}$ is an integral domain. The commutativity computation is the displayed example.

Invertibility

Theorem. $T_a$ is invertible if and only if $a$ is a unit of $\mathcal{A}$, that is, if and only if $a(1)\ne0$, and then $$ T_a^{-1}=\alpha L_{a^{-1}}=T_{\alpha(a^{-1})}. $$

Proof. $\alpha$ is invertible and $L_a$ is invertible exactly for $a$ a unit; the inverse of the composite $L_a\alpha$ is $\alpha^{-1}L_a^{-1}=\alpha L_{a^{-1}}$, and $\alpha L_{a^{-1}}=L_{\alpha(a^{-1})}\alpha=T_{\alpha(a^{-1})}$.

The Adjoint and the Conditions

The adjoint

Theorem (the signed adjoint). With respect to the form of the category, $$ (T_a)^* = \alpha\,\Theta_a , $$ and explicitly $$ (T_a)^*(g)(k)=\lambda(k)\sum_{m\ge1}\overline{a(m)}\,g(mk). $$ One has $T_a^*T_a=\alpha\,\Theta_aL_a\,\alpha$ and $T_aT_a^*=L_a\Theta_aL_{\alpha(a)}$; in particular the two products are unequal in general although each is a product of a multiplication and its transpose.

Proof. $\alpha^*=\alpha$ and $L_c^*=\Theta_c$ give $(L_a\alpha)^*=\alpha L_a^*=\alpha\Theta_a$. The explicit formula is the definition of $\Theta_a$ multiplied by $\lambda$. The two products are computed from the first display; the product $L_a\Theta_aL_{\alpha(a)}$ is not equal to $L_{\alpha(a)}\Theta_aL_a$ in general.

Corollary (comparison with the ring formula). In the noncommutative ring of the earlier article the adjoint of the signed left multiplication by $a$ is the signed left multiplication by $\delta(a)=\sigma\alpha(a)$; here the adjoint is the transposed multiplication $\Theta_a$ composed with the grade involution, and it is not a signed left multiplication unless $\alpha(a)$ is a unit. The discrepancy is the commutativity of the algebra together with the failure of the two structures (involution on the elements, adjoint on the operators) to coincide except in the degenerate augmentation form.

Self-adjointness, isometry and unitarity

Theorem. The signed left multiplication $T_a$ satisfies the following. $$ T_a\ \text{is self-adjoint} \iff a=\gamma\varepsilon \text{ with } \gamma \text{ real},\qquad T_a\ \text{is unitary} \iff a=\gamma\varepsilon \text{ with } |\gamma|=1, $$ and $T_a$ is an isometry if and only if $\Theta_aL_a=\mathrm{id}$, which holds for $a=\gamma\delta_p$ with $|\gamma|=1$ and more generally whenever the isometry criterion of the signed sandwiches holds.

Proof. Self-adjointness is $\alpha\Theta_a=L_a\alpha$. Applying both sides to $\delta_1$: the left side is $\alpha(\overline{a(1)}\delta_1)=\overline{a(1)}\delta_1$, and the right side is $a$; hence $a$ is supported at $1$, say $a=\gamma\delta_1$, and the equation becomes $\overline{\gamma}\delta_1=\gamma\lambda\delta_1=\gamma\delta_1$, so $\gamma$ is real. Unitarity is $T_a^*T_a=\mathrm{id}$ and $T_aT_a^*=\mathrm{id}$; the first is $\alpha\Theta_aL_a\alpha=\mathrm{id}$, that is $\Theta_aL_a=\mathrm{id}$, and the second then forces $L_a\Theta_a=\mathrm{id}$; as for the sandwich, the unitary case is $a=\gamma\varepsilon$ with $|\gamma|=1$. The isometry statement is the criterion $\Theta_aL_a=\mathrm{id}$, verified for $a=\gamma\delta_p$ with $|\gamma|=1$ by the computation of the shifts.

Corollary. The only self-adjoint signed left multiplications are the real scalar multiples of $\alpha$; the only unitary ones are the unitary scalar multiples of $\alpha$; the signed left multiplication by $\delta_p$ is a nonunitary isometry with cokernel the functions vanishing on the multiples of $p$.

Worked Examples

Example ($a=\varepsilon$). $T_\varepsilon=\alpha$, the grade involution of Reflections as Signed Two-Sided Operators on the Algebra of Arithmetic Functions; it is unitary and self-adjoint.

Example ($a=\delta_p$). $T_{\delta_p}(f)=\delta_p*\alpha(f)$, with $(T_{\delta_p}f)(n)=[p\mid n]\lambda(n/p)f(n/p)$; it is an isometry with cokernel the functions vanishing on the multiples of $p$, and $(T_{\delta_p})^*=\alpha\Theta_{\delta_p}=\alpha S_p$, where $S_p$ is the division of The Shift Operator on the Coefficients.

Example ($a=\mathbf 1$). $T_{\mathbf 1}f=\mathbf 1*\alpha(f)$; the function $\mathbf 1$ has infinite support, so $T_{\mathbf 1}$ is not self-adjoint, not an isometry and not unitary, and it is unbounded on the dense finitely supported subspace.

Example ($a=\mu$). $T_\mu=L_\mu\alpha=\alpha L_{\lambda\mu}$; since $\lambda\mu$ is completely multiplicative with values in $\{-1,0,1\}$, $L_{\lambda\mu}$ is a contraction, and $T_\mu^*T_\mu=\alpha\Theta_\mu L_\mu\alpha$ is the projection onto the squarefree integers up to the sign, so $T_\mu$ is not an isometry.

Failure of the Degenerate Cases

The signed left multiplication fails to be a genuinely signed operator over the commutative algebra in four ways. First, $T_a=L_a\alpha=\alpha L_{\alpha(a)}$ is a composite of the fixed automorphism $\alpha$ with an unsigned multiplication, so the sign contains no information beyond the value of $\alpha$. Second, the adjoint is $\alpha\Theta_a$, not a signed left multiplication, so the signed adjoint formula of the ring theory does not transfer. Third, the products of two signed left multiplications are unsigned, $T_aT_b=L_{a*\alpha(b)}$, and the family is not commutative; the twisted product $\star$ is the correct product, and it is isomorphic to the original one through $\alpha$. Fourth, self-adjointness and unitarity hold only for the scalar multiples of $\alpha$, so the class of distinguished signed left multiplications is as small as possible. These are the boundary cases of the general theory, collected because they are the reason the signed family over a commutative algebra is a family of examples rather than a family of structures.

Summary

For $a$ in the algebra of arithmetic functions the signed left multiplication is $T_a(f)=a*\alpha(f)=L_a\alpha=\alpha L_{\alpha(a)}$, with $\alpha T_a=L_{\alpha(a)}$ and $T_a\alpha=L_a$; the twisted product $a\star b=a*\alpha(b)$ is associative, $\alpha$ is an isomorphism $(\mathcal{A},\star)\to(\mathcal{A},*)$, and $T_aT_b=L_{a*\alpha(b)}$, so products of signed left multiplications are unsigned and the family is not commutative. The operator is invertible exactly for $a$ a unit, with $T_a^{-1}=\alpha L_{a^{-1}}$; its adjoint for the coefficient form is $(T_a)^*=\alpha\Theta_a$, with $(T_a)^*g(k)=\lambda(k)\sum_m\overline{a(m)}g(mk)$; it is self-adjoint exactly for $a=\gamma\varepsilon$ with $\gamma$ real, unitary exactly for $a=\gamma\varepsilon$ with $|\gamma|=1$, and an isometry exactly when $\Theta_aL_a=\mathrm{id}$, which holds for $a=\gamma\delta_p$ with $|\gamma|=1$. The operator is injective for every nonzero $a$ because the algebra is an integral domain, and the signed adjoint formula of the noncommutative ring theory fails.

Summary of Notation

Symbol	Meaning
$T_a(f)=a*\alpha(f)$	Signed left multiplication
$T_a=L_a\alpha=\alpha L_{\alpha(a)}$	The two readings
$a\star b=a*\alpha(b)$	The twisted product
$T_aT_b=L_{a\star b}$	Product law
$T_a^{-1}=\alpha L_{a^{-1}}$	Inverse for a unit
$(T_a)^*=\alpha\Theta_a$	The signed adjoint
$(T_a)^*g(k)=\lambda(k)\sum_m\overline{a(m)}g(mk)$	Explicit adjoint
$T_a$ self-adjoint $\iff$ $a=\gamma\varepsilon$, $\gamma$ real	Self-adjointness
$T_a$ unitary $\iff$ $a=\gamma\varepsilon$, $\|\gamma\|=1$	Unitarity
$\Theta_aL_a=\mathrm{id}$	Isometry condition