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Let $G$ be a simply connected compact Lie group and $T$ its maximal torus with inclusion $i:T \hookrightarrow G$.

By simply connectedness of the group $G$ and asphericity of the torus $T$, the induced map $i_* : \pi_*(T) \to \pi_*(G)$ is trivial. Moreover, the maximal torus of $S^3$ or of $Spin(4) \cong S^3 \times S^3$ has a null-homotopic inclusion into the respective group.

Therefore, I am wondering if the above indicates that this is true in greater generality, i.e. is the inclusion $i:T\hookrightarrow G$ of the maximal torus in a simply-connected Lie group null-homotopic?

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Yes.

Write $T=T_1\times\dots\times T_k$, with each $T_i$ isomorphic to the circle group. Since $T_i$ is a circle and $G$ is simply connected, there exists a homotopy $(f_i(t,x))$, with $f_i(0,x)=x$ and $f_i(1,x)=1$ for $x\in T_i$.

Then for $x=(x_1,\dots,x_k)\in T$, write $f(t,x)=\prod_{i=1}^kf_i(t,x)$. Then $f(0,x)=x$ and $f(1,x)=1$.

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  • $\begingroup$ Wouldn't this argument prove that any map $T^3\to S^3$ is null-homotopic, including the obvious degree one map? $\endgroup$
    – Igor Belegradek
    Commented Oct 10, 2023 at 13:55
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    $\begingroup$ @IgorBelegradek No. Indeed it uses the fact that $(x_1,\dots,x_k)$ equals the product $x_1\dots x_k$, which is needed to get $f(0,x)=x$. $\endgroup$
    – YCor
    Commented Oct 10, 2023 at 14:05

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