This is somehow very natural question so I'm sure that the answer should be well known: Whitney theorem states that each (say paracompact) $n$-dimensional manifold could be embedded in $\mathbb{R}^{2n}$-but very often $2n$ is not necessesary, one could find smaller $m$ such that $n$ dimensional manifold could be embedded into $\mathbb{R}^{m}$. What are the minimal numbers $m$ such that the classical Lie groups such as $SU(n)$ or $SO(n)$ can be embedded in $\mathbb{R}^m$? What about the torus $\mathbb{T}^n$-could be always embedded in $\mathbb{R}^{n+1}$?
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For the torus, the answer is positive, see: Embedding of n-Torus in $\mathbb{R}^{n+1}$ via implicit function $T^n=f^{-1}(0)$
For such compact groups if the dimension is not a power of two ( I'm pretty sure $\frac {n(n-1)} 2 \not= 2^k$ for all $n,k$) they can be embedded in $\mathbb R ^{2n-1}$ and that's the minimal n for $SO(3)$, but for $SU(2)$ is 4. Since $SU(n)$ is 2-connected it can be embedded in $\mathbb R ^{2n-2}$. Since any simplify connected lie group is also 2 connected the result can be improved to $\mathbb R ^{2n-2}$ for those if their dimension is bigger than 7.
See the Wikipedia article for the withney embedding theorem https://en.m.wikipedia.org/wiki/Whitney_embedding_theorem
k76u4vkweek547v7
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Of course this is not a complete answer but I think being a lie group doesn't help to much in this aspect – k76u4vkweek547v7 Jan 31 '16 at 17:21
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1Being a Lie group helps a lot; it ensures that the tangent bundle is trivial, which removes a possible obstruction. So e.g. a connected Lie group of dimension $d$ immerses into $\mathbb{R}^{d+1}$ by the Hirsch-Smale theorem. – Qiaochu Yuan Jan 31 '16 at 18:58
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Do you know criteria to improve immersion to embedding? – k76u4vkweek547v7 Jan 31 '16 at 19:24
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$SU(2)$ and $SO(3)$ are isomorphic! – Natha Jan 08 '17 at 15:51