Let $G=\mathrm{GL}_2$ and $s=\left(\begin{smallmatrix} a & \\ & b \end{smallmatrix}\right)$ be semisimple and non-regular in $G(\mathbb{F}_q)=\mathrm{GL}_2(\mathbb{F}_q)$ (i.e. $a\neq b$ and $ab\neq 0$). In their 1976 paper, Deligne-Lusztig gave a formula for computing $\chi(s)$ using Deligne-Lusztig characters (where $\chi$ is an irreducible character of $\mathrm{GL}_2(\mathbb{F}_q)$):
Corollary 7.6 of Deligne-Lusztig: $$ \chi(s) = \frac{1}{\mathrm{St}_G(s)}\sum_{T\ni s} \sum_{\theta\in \widehat{T}} \theta(s) (-1)^{\sigma(G)-\sigma(T)}\langle \chi, R_T^G(\theta)\rangle $$
where
- $\mathrm{St}_G(s)$ is the Steinberg representation of $G=\mathrm{GL}_2$, which takes the value $q$ on semisimple non-regular $s$,
- the first sum is over all tori containing $s$, so we sum over the maximal split torus $T_1 = \{\left(\begin{smallmatrix} x & \\ & y \end{smallmatrix}\right)\ |\ x,y\in\mathbb{F}_q,\ xy\neq 0\}$ and the anisotropic torus $T_2 = \{\left(\begin{smallmatrix} x & \varepsilon y \\ y & x \end{smallmatrix}\right)\ |\ x,y\in\mathbb{F}_q,\ x^2-\varepsilon y^2 \neq 0\}$,
- $\sigma(G)$ is the $\mathbb{F}_q$-rank of $G$, so $\sigma(G)=2$, $\sigma(T_1)=2$ and $\sigma(T_1)=1$.
I want to calculate the above for any semisimple non-regular $s$, but I'm still stuck on the case of $s=1$ and $\chi=1$. Here's my calculation:
\begin{align*} \chi(1) &= \frac{1}{q}\bigg[\sum_{\theta\in \widehat{T_1}} (-1)^{\sigma(G)-\sigma(T_1)}\langle \chi, R_{T_1}^G(\theta)\rangle + \sum_{\theta\in \widehat{T_2}} (-1)^{\sigma(G)-\sigma(T_2)}\langle \chi, R_{T_2}^G(\theta)\rangle\bigg] \\ &= \frac{1}{q}\bigg[\sum_{\theta\in \widehat{T_1}} \langle \chi, R_{T_1}^G(\theta)\rangle - \sum_{\theta\in \widehat{T_2}} \langle \chi, R_{T_2}^G(\theta)\rangle\bigg]. \tag{1}\label{1} \end{align*}
Then $$ \sum_{\theta\in \widehat{T_1}} \langle \chi, R_{T_1}^G(\theta)\rangle = \sum_{\alpha,\beta\in\widehat{\mathbb{F}_q^\times}} \langle \chi,R_{T_1}^G(\theta_{\alpha,\beta})\rangle = \sum_{\substack{\alpha,\beta\in\widehat{\mathbb{F}_q^\times} \\ \alpha\neq\beta}} \langle \chi,R_{T_1}^G(\theta_{\alpha,\beta})\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \langle \chi,R_{T_1}^G(\theta_{\alpha,\alpha})\rangle, $$ where $\theta_{\alpha,\beta}\left(\begin{smallmatrix} x & \\ & y \end{smallmatrix}\right) := \alpha(x)\beta(y)$. Now if $\alpha\neq\beta$ then $R_{T_1}^G(\theta_{\alpha,\beta}) = W_{\alpha,\beta}$, the principle series representation (in the notation of Fulton and Harris, page 70), and $R_{T_1}^G(\theta_{\alpha,\alpha}) = U_\alpha(R_{T_1}^G(1)) = U_\alpha(1+\mathrm{St}_G) = U_\alpha + V_\alpha$. Here, $U_\alpha$ is the determinant rep obtained from $\alpha$ and $V_\alpha = U_\alpha\otimes \mathrm{St}$.
Then $$ \sum_{\substack{\alpha,\beta\in\widehat{\mathbb{F}_q^\times} \\ \alpha\neq\beta}} \langle \chi,R_{T_1}^G(\theta_{\alpha,\beta})\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \langle \chi,R_{T_1}^G(\theta_{\alpha,\alpha})\rangle = \sum_{\substack{\alpha,\beta\in\widehat{\mathbb{F}_q^\times} \\ \alpha\neq\beta}} \langle \chi,W_{\alpha,\beta}\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \langle \chi,U_\alpha+V_\alpha\rangle, $$ but $\chi=1=U_1$ so the above reduces to $1$ (obtained from the second sum when $\alpha=1$).
On the other hand, we have $$ \sum_{\theta\in \widehat{T_2}} \langle \chi, R_{T_2}^G(\theta)\rangle = \sum_{\varphi\in \widehat{\mathbb{F}_{q^2}^\times}} \langle \chi, R_{T_2}^G(\varphi)\rangle. $$ Now if $\varphi\in\widehat{\mathbb{F}_{q^2}^\times}$ is indecomposable (i.e. does not equal $\alpha\circ\mathrm{Norm}$ for some $\alpha\in\widehat{\mathbb{F}_q^\times}$) then $R_{T_2}^G(\varphi)=-X_\varphi$, where $X_\varphi$ is the cuspidal rep associated to $\varphi$ (in the notation of Fulton and Harris). Otherwise, $\varphi=\alpha\circ\mathrm{Norm}$ and $R_{T_2}^G(\varphi) = R_{T_2}^G(\alpha\circ\mathrm{Norm}) = U_\alpha$.
Then \begin{align*} \sum_{\varphi\in \widehat{\mathbb{F}_{q^2}^\times}} \langle \chi, R_{T_2}^G(\varphi)\rangle &= \sum_{\substack{\varphi\in\widehat{\mathbb{F}_{q^2}^\times} \\ \varphi\ \text{indecomposable}}} \langle \chi, R_{T_2}^G(\varphi)\rangle + \sum_{\substack{\varphi\in\widehat{\mathbb{F}_{q^2}^\times} \\ \varphi\ \text{decomposable}}} \langle \chi, R_{T_2}^G(\varphi)\rangle \\ &= \sum_{\substack{\varphi\in\widehat{\mathbb{F}_{q^2}^\times} \\ \varphi\ \text{indecomposable}}} \langle \chi, -X_\varphi\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \langle \chi, U_\alpha\rangle \\ &= 1. \end{align*} This means that $\chi(1)=\frac{1}{q}(1-1)=0$ when it should be $1$.
I believe my mistake is in my understanding of $T_2$ and it's irreducible characters, since my calculations of $\chi(s)$ for regular $s$ and irreducible $\chi$ are all correct (verified against the character table of $\mathrm{GL}_2(\mathbb{F}_q)$), and you don't need to consider $T_2$ for regular $s$.
Edit: I've made a mistake in \eqref{1}, I removed the $\theta(s)$ coefficients without justification. It should be $$ \chi(s) = \frac{1}{q}\bigg[\sum_{\theta\in \widehat{T_1}} \theta(s)\langle \chi, R_{T_1}^G(\theta)\rangle - \sum_{\theta\in \widehat{T_2}} \theta(s)\langle \chi, R_{T_2}^G(\theta)\rangle\bigg]. $$ I'm going to let $\chi=U_\gamma$ be the determinant representation associated to $\gamma$, and $s=\left(\begin{smallmatrix} a & \\ & a \end{smallmatrix}\right)$. For the first term, we have $$ \sum_{\theta\in \widehat{T_1}} \theta(s)\langle U_\gamma, R_{T_1}^G(\theta)\rangle = \sum_{\alpha,\beta\in\widehat{\mathbb{F}_q^\times}} \theta_{\alpha,\beta}(s)\langle U_\gamma,R_T^G(\theta_{\alpha,\beta})\rangle = \sum_{\substack{\alpha,\beta\in\widehat{\mathbb{F}_q^\times}}\\ \alpha\neq\beta} \theta_{\alpha,\beta}(s)\langle U_\gamma,R_T^G(\theta_{\alpha,\beta})\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \theta_{\alpha,\alpha}(s)\langle U_\gamma,R_T^G(\theta_{\alpha,\alpha})\rangle = \sum_{\substack{\alpha,\beta\in\widehat{\mathbb{F}_q^\times}}\\ \alpha\neq\beta} \theta_{\alpha,\beta}(s)\langle U_\gamma,W_{\alpha,\beta}\rangle + \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \theta_{\alpha,\alpha}(s)\langle U_\gamma,U_\alpha+V_\alpha\rangle = 0 + \theta_{\gamma,\gamma}(s) = \gamma(a^2). $$ Then, for the second term, we have $$ \sum_{\theta\in \widehat{T_2}} \theta(s)\langle U_\gamma, R_{T_2}^G(\theta)\rangle = \sum_{\substack{\theta\in \widehat{T_2} \\ \theta\neq\theta^q}} \theta(s)\langle U_\gamma, R_{T_2}^G(\theta)\rangle + \sum_{\substack{\theta\in \widehat{T_2} \\ \theta=\theta^q}} \theta(s)\langle U_\gamma, R_{T_2}^G(\theta)\rangle $$ For the first term, $R_{T_2}^G(\theta)=-X_\theta$ (the cuspidal rep associated to $\theta$) when $\theta\neq\theta^q$, so this term disappears.
However, now my problem is calculating the second term. I know that if $\theta=\theta^q$ then $\theta$, when considered as a character of $\mathbb{F}_{q^2}^\times(\cong T_2)$, is of the form $\alpha\circ\mathrm{Norm}$ for some $\alpha\in\widehat{\mathbb{F}_q}$, and $R_{T_2}^G(\theta)=U_\alpha(R_{T_2}^G(1_{T_2})) = U_\alpha(1-\mathrm{St})=U_\alpha-V_\alpha$. Then we use the association $\left(\begin{smallmatrix} a & \varepsilon b \\ b & a \end{smallmatrix}\right)\leftrightarrow a+b\sqrt{\varepsilon}$ to associate $\left(\begin{smallmatrix} a & 0 \\ 0 & a \end{smallmatrix}\right)\leftrightarrow a+0\sqrt{\varepsilon}$ and compute $$ \sum_{\substack{\theta\in \widehat{T_2} \\ \theta=\theta^q}} \theta(s)\langle U_\gamma, R_{T_2}^G(\theta)\rangle = \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} (\alpha\circ\mathrm{Norm})(s)\langle U_\gamma, U_\alpha-V_\alpha\rangle = \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \alpha(\mathrm{Norm}(a+0\varepsilon))\langle U_\gamma, U_\alpha-V_\alpha\rangle = \sum_{\alpha\in\widehat{\mathbb{F}_q^\times}} \alpha(a^2)\langle U_\gamma, U_\alpha-V_\alpha\rangle = \gamma(a^2) $$ But then I get $$ \chi_{U_\gamma}(s)=\chi(s) = \frac{1}{q}\bigg[\sum_{\theta\in \widehat{T_1}} \theta(s)\langle \chi, R_{T_1}^G(\theta)\rangle - \sum_{\theta\in \widehat{T_2}} \theta(s)\langle \chi, R_{T_2}^G(\theta)\rangle\bigg] = \frac{1}{q}\bigg[(0+\gamma(a^2)) - (0+\gamma(a^2))\bigg] =0. $$
Edit 2: I have found an interesting statement that may be exactly what's needed to resolve this calculation. From Geck-Malle (page 114):
The character formula immediately shows that, if $z\in Z(G)^F$ then we have $R_{T}^G(\theta)(z) = \theta(z)R_T^G(\theta)(1)$. Something stronger is true.
Prop 2.2.20 Let $T\subseteq G$ be an $F$-stable maximal torus and $\theta\in\widehat{T^F}$. Let $\chi\in\widehat{G^F}$ be such that $\langle \chi,R_T^G(\theta),\chi\rangle\neq 0$. Then $\chi(z)=\theta(z)\chi(1)$ for all $z\in Z(G)^F$.
Here, $G^F=G(\mathbb{F}_q)$, $T^F=T(\mathbb{F}_q)$ and $Z(G)^F=Z(G(\mathbb{F}_q))$.
