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In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups.

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n+1,\mathbb C)$, $G(\mathbb R)=Spin(n,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n)$ is simply connected, so $Spin(n,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.

In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups. [Real forms of exceptional groups may be identified by their maximal compact subgroups; $E_6(F_4)$ is the real form of $E_6$ with maximal compact subgroup of type $F_4$, known in another classification as $E_{6(-20)}$; $F_4(B_4)=F_{4(-20)}$ has maximal compact $B_4$.]

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n+1,\mathbb C)$, $G(\mathbb R)=Spin(n,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n)$ is simply connected, so $Spin(n,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.

In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups.

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n,\mathbb C)$, $G(\mathbb R)=Spin(n-1,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n-1)$ is simply connected, so $Spin(n-1,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.

In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups.

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n+1,\mathbb C)$, $G(\mathbb R)=Spin(n,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n)$ is simply connected, so $Spin(n,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.

In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups.

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n,\mathbb C)$, $G(\mathbb R)=Spin(n-1,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n-1)$ is simply connected, so $Spin(n-1,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$.

In fact "most" simple groups have such covers: if and only if there is a long real root. There is a long and rich history of this topic, going back at least to a paper by Jacque Tits from 1967 (as well as some older work which appeared only in Russian). For a uniform (case-free) treatment see Nonlinear covers of real groups, IMRN 2004, Math Reviews 2112326. Also see the references in the paper and in the Mathscinet review. A good basic reference is Onischik and Vinberg, Lie Groups and Algebraic Groups; the answer can be read off from Table 10 in the Reference Chapter.

Suppose $G(\mathbb C)$ is (connected) simple, complex, simply connected, and $G(\mathbb R)$ is a real form. Then $G(\mathbb R)$ has a covering group with no faithful finite dimensional representation if and only if $G(\mathbb R)$ is not simply connected. Let $K(\mathbb R)$ be a maximal compact subgroup, with complexification $K(\mathbb C)$. Then $G(\mathbb R), K(\mathbb R)$ and $K(\mathbb C)$ all have the same fundamental group. So the question becomes: is $\pi_1(K(\mathbb C))=1$?

Examples where $G(\mathbb R)$ is simply connected are rare. Here is the complete list: compact, complex, $SL(n,\mathbb H)$, $Spin(n,1)$ $(n\ge 3)$, $Sp(p,q)$, $E_6(F_4)$ and $F_4(B_4)$. (By complex I mean a complex group, viewed as a real Lie group). If G is simply laced then $\pi_1(G(\mathbb R))=1$ if and only if $G(\mathbb R)$ has only one conjugacy class of Cartan subgroups.

Example: $G(\mathbb C)=SL(n,\mathbb C)$, $G(\mathbb R)=SL(n,\mathbb R)$, $K(\mathbb R)=SO(n,\mathbb R)$, $K(\mathbb C)=SO(n,\mathbb C)$. If $n\ge 3$ then $SO(n,\mathbb C)$ has a two-fold cover $Spin(n,\mathbb C)$, and the universal cover of $SL(n,\mathbb R)$ is two-fold, with maximal compact subgroup $Spin(n)$. If $n=2$, $SO(2,\mathbb C)=\mathbb C^*$, $\pi_1=\mathbb Z$, and the universal cover of $SL(2,\mathbb R)$ is infinite.

Example: $G(\mathbb C)=Spin(n,\mathbb C)$, $G(\mathbb R)=Spin(n-1,1)$ ($n\ge 3$), $K(\mathbb R)=Spin(n-1)$ is simply connected, so $Spin(n-1,1)$ is simply connected. For example $Spin(3,1)\simeq SL(2,\mathbb C)$ is simply connected.