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This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type defined of a large field.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot r^2 \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probably show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).

This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type defined over a large field.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot r^2 \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probably show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).

This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type defined of a large field.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot r^2 \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probabibly show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).

This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type defined of a large field.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot r^2 \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probably show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).

This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probabibly show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).

This answer shows that one cannot find $(G,\rho)$ as required if $G$ is supposed to be group of Lie type defined of a large field.

Let $G$ be a group of Lie type defined over a field with $q$ elements. Let $\rho$ be an irreducible representation of $G$ which is not $1$-dimensional. Gluck (D. Gluck. Sharper character value estimates for groups of Lie type, J. Algebra 174 (1995), 229-266.) proved that there exists a constant (independent of $G$ and $\rho$) so that

$$|{\rm tr}(\rho(g))| \leq C \cdot \dim(\rho) \cdot q^{-1/2},$$ whenever $g$ is non-trival.

Let's assume that $\rho$ is unitary and denote by $\|.\|$ the spectral norm. Now, let $f \in {\mathbb C}[G]$, $f= \sum_{i=1}^r c_i g_i$ with $g_i \neq g_j$ for $i \neq j$. We get

$${\rm tr}(\rho(f)^*\rho(f)) \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2,$$ and \begin{eqnarray*} {\rm tr}(\rho(f)^*\rho(f)) &=& \sum_{i,j=1}^r c_i \bar{c_j} \cdot {\rm tr}(\rho(g_ig_j^{-1}))\\ &\geq & \dim(\rho) \cdot \sum_{i=1}^r |c_i|^2 - C \cdot \dim(\rho) \cdot q^{-1/2} \cdot \sum_{i \neq j} |c_i \bar{c_j}| \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \left(\sum_{i=1}^r |c_i| \right)^2 \\ &\geq & \dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2, \end{eqnarray*} where I used Gluck's result and the basic estimate $$\sum_{i=1}^r |c_i| \leq r^{1/2} \cdot \left(\sum_{i=1}^r |c_i|^2 \right)^{1/2}.$$

Combining the two bounds for ${\rm tr}(\rho(f)^*\rho(f))$ we get $$\dim(\rho) \cdot \left(1/r - C \cdot q^{-1/2} \right) \cdot \|\rho(f)\|^2 \leq {\rm rank}(\rho(f)) \cdot \|\rho(f)\|^2.$$ and hence $$\frac{\dim(\rho)}{{\rm rank}(\rho(f))} \leq \frac1{1/r - C \cdot q^{-1/2}} \leq r + 2 C \cdot r^2 \cdot q^{-1/2},$$ for $q$ large.

There are similar character estimates for the symmetric group, comparing an arbitrary character with the character that counts fixed points of the permutation. These estimates probabibly show that the symmetric groups cannot work either (at least not in a way that is essentially better than for the standard representation).