Let $G$ be a simply connected semisimple group over a perfect field $k$ (at the moment I am interested in the case $k=\mathbb R$). Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$: there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that $G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$. Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$. Let $$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ denote the connecting map from the cohomology exact sequence $$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ induced by the short exact sequence $$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$ By definition, the Tits class of $G$ is $$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$ see The Book of Involutions, (31.7). Note that $Z_{\rm qs}=Z:=Z(G)$.

Question 1. Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$?

For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split) we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$. From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$. I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; in all the other cases we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$.

However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$. I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$, but I need an explicit formula for the cases ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$. Thus Question 1 reduces to the following question:

Question 2. What are the Tits classes for the simple ${\mathbb R}$-groups ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ?

In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$. Such a description is given, for example, in Brian Conrad's cheat sheet.

Let $G$ be a simply connected semisimple group over a perfect field $k$ (at the moment I am interested in the case $k=\mathbb R$). Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$: there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that $G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$. Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$. Let $$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ denote the connecting map from the cohomology exact sequence $$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ induced by the short exact sequence $$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$ By definition, the Tits class of $G$ is $$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$ see The Book of Involutions, (31.7). Note that $Z_{\rm qs}=Z:=Z(G)$.

Question 1. Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$?

For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split) we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$. From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$. I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; for non-strong inner forms we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$.

However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$. I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$, but I need an explicit formula for the cases ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$. Thus Question 1 reduces to the following question:

Question 2. What are the Tits classes for the simple ${\mathbb R}$-groups ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ?

In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$. Such a description is given, for example, in Brian Conrad's cheat sheet.

Let $G$ be a semisimple group over a perfect field (at the moment I am interested in the case $k=\mathbb R$). Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$: there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that $G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$. Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$. Let $$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ denote the connecting map from the cohomology exact sequence $$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ induced by the short exact sequence $$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$ By definition, the Tits class of $G$ is $$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$ see The Book of Involutions, (31.7). Note that $Z_{\rm qs}=Z:=Z(G)$.

Question 1. Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$?

For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split) we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$. From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$. I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; in all the other cases we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$.

However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$. I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$, but I need an explicit formula for the cases ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$. Thus Question 1 reduces to the following question:

Question 2. What are the Tits classes for the simple ${\mathbb R}$-groups ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ?

In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$. Such a description is given, for example, in Brian Conrad's cheat sheet.

Let $G$ be a simply connected semisimple group over a perfect field $k$ (at the moment I am interested in the case $k=\mathbb R$). Then $G$ is an inner form of a quasi-split $k$-group $G_{\rm qs}$: there exists a quasi-split form $G_{\rm qs}$ of $G$ and a 1-cocycle $c\in Z^1(k,\overline{G}_{\rm qs})$ such that $G=\,_c G_{\rm qs}$, the inner twist of $G_{\rm qs}$ by the cocycle $c$, where $\overline{G}_{\rm qs}=G_{\rm qs}/Z_{\rm qs}$ and $Z_{\rm qs}=Z(G_{\rm qs})$. Let $\xi=[c]\in H^1(k,\overline{G}_{\rm qs})$, the cohomology class of the cocycle $c$. Let $$\Delta\colon H^1(k,\overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ denote the connecting map from the cohomology exact sequence $$H^1(k,Z_{\rm qs})\to H^1(k,G_{\rm qs})\to H^1(k, \overline{G}_{\rm qs})\to H^2(k,Z_{\rm qs})$$ induced by the short exact sequence $$ 1\to Z_{\rm qs}\to G_{\rm qs}\to \overline{G}_{\rm qs}\to 1.$$ By definition, the Tits class of $G$ is $$t_G=\Delta(\xi)\in H^2(k,Z_{\rm qs}),$$ see The Book of Involutions, (31.7). Note that $Z_{\rm qs}=Z:=Z(G)$.

Question 1. Is there anywhere in the literature or in the Internet a table of $t_G$ for all simply connected absolutely simple ${\mathbb R}$-groups $G$?

For all connected Dynkin diagram except for ${\mathsf D}_{2m}$ (and also for ${\mathsf D}_{2m}$ when $G_{\rm qs}$ is not split) we have either $H^2({\mathbb R},Z_{\rm qs})=0$ or $H^2({\mathbb R},Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}$. From the cohomology exact sequence we see that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, $\xi$ comes from $H^1({\mathbb R}, G)$. I have a table of strong inner forms of $G_{\rm qs}$, and so I know when $t_G=0$; in all the other cases we have $t_G\neq 0$, hence $t_G=1+2{\mathbb Z}\in {\mathbb Z}/2{\mathbb Z}$.

However, for ${\mathsf D}_{2m}$, in the case when $G_{\rm qs}$ is split, namely $G_{\rm qs}\simeq {\bf Spin}(2m,2m)$, we have $H^2({\mathbb R}, Z_{\rm qs})={\mathbb Z}/2{\mathbb Z}\times{\mathbb Z}/2{\mathbb Z}$. I know that $t_G=0$ if and only if $G$ is a strong inner form of $G_{\rm qs}$, that is, if $G\simeq{\bf Spin}(2m+4q, 2m-4q)$, but I need an explicit formula for the cases ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and the quaternionic form ${\bf Spin}^*(4m)$. Thus Question 1 reduces to the following question:

Question 2. What are the Tits classes for the simple ${\mathbb R}$-groups ${\bf Spin}(2m+4q+2, 2m-4q-2)$ and ${\bf Spin}^*(4m)$ ?

In order to formulate an answer to Question 2, one needs an explicit description of the center of the split ${\mathbb R}$-group ${\bf Spin}(2m,2m)$. Such a description is given, for example, in Brian Conrad's cheat sheet.