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Mikael de la Salle
Mikael de la Salle
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Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.

In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$.

The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).

One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.

For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf S$$F_S^{ab}=\mathbf Z^S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation  $z^{(1)}=z$ and $z^{(-1)}=\bar z$.

Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write $$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$ QED

Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the Banachlinear space space spanned by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated. When $u_i=g_i$ (resp $v_i=g_i'$) in $C^*(G)$ (resp. $C^*(G')$), this implies that $G$ and $G'$ are isomorphic.

Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.

In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$.

The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).

One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.

For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation$z^{(1)}=z$ and $z^{(-1)}=\bar z$.

Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write $$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$ QED

Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the Banach space space by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated.

Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.

In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$.

The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).

One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.

For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf Z^S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation  $z^{(1)}=z$ and $z^{(-1)}=\bar z$.

Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write $$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$ QED

Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the linear space space spanned by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated. When $u_i=g_i$ (resp $v_i=g_i'$) in $C^*(G)$ (resp. $C^*(G')$), this implies that $G$ and $G'$ are isomorphic.

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Mikael de la Salle
Mikael de la Salle
  • 9.5k
  • 2
  • 31
  • 46

Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.

In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$.

The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).

One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.

For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation$z^{(1)}=z$ and $z^{(-1)}=\bar z$.

Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write $$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$ QED

Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the Banach space space by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated.