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Yang
Yang
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In the construction of $K_0^{G\times\mathbb{G_m}}(St)\cong H_{aff}$$K_0^{G\times\mathbb{G_m}}(St)\cong H_{\mathrm{aff}}$ the class of $\Delta_* L_\lambda$ is mapped to $e^{-\lambda}$, so the action of $e^{\lambda}\in R(T)\subset H_{aff}$$e^{\lambda}\in R(T)\subset H_{\mathrm{aff}}$ on $R(T)$ should be multiplication by $e^{-\lambda}$. Then everything will be good.

In the construction of $K_0^{G\times\mathbb{G_m}}(St)\cong H_{aff}$ the class of $\Delta_* L_\lambda$ is mapped to $e^{-\lambda}$, so the action of $e^{\lambda}\in R(T)\subset H_{aff}$ on $R(T)$ should be multiplication by $e^{-\lambda}$. Then everything will be good.

In the construction of $K_0^{G\times\mathbb{G_m}}(St)\cong H_{\mathrm{aff}}$ the class of $\Delta_* L_\lambda$ is mapped to $e^{-\lambda}$, so the action of $e^{\lambda}\in R(T)\subset H_{\mathrm{aff}}$ on $R(T)$ should be multiplication by $e^{-\lambda}$. Then everything will be good.

Source Link
Yang
Yang
  • 71
  • 3

In the construction of $K_0^{G\times\mathbb{G_m}}(St)\cong H_{aff}$ the class of $\Delta_* L_\lambda$ is mapped to $e^{-\lambda}$, so the action of $e^{\lambda}\in R(T)\subset H_{aff}$ on $R(T)$ should be multiplication by $e^{-\lambda}$. Then everything will be good.