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I've seen that there are different groups like $SU(2)$ and $SO(3)$ with isomorphic Lie algebra. I know that putting the elements of the linear algebra in the exponential function gives the group elements, in this case I don't understand how it's possible that two isomorphic linear algebras gives different groups

Does it mean that when I put two isomorphic group in the exponential function I can obtain two non isomorphic groups?

SimoBartz
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    Not all the properties of the set of matrices you get by the exponential are coded in the Lie algebra alone. In many interesting cases (such as in your example) the resulting Lie groups will share the same universal cover, and are closely related. – Jyrki Lahtonen May 13 '20 at 10:35
  • Does it mean that when I put two isomorphic group in the exponential function I can obtain two non isomorphic groups? – SimoBartz May 13 '20 at 10:39
  • Basically you put the same Lie algebra into two different exponential functions. With freely chosen Lie algebras, even though isomorphic, funnier things can happen. Consider what happens with 1-dimensional (real) Lie algebras spanned by $$A=\pmatrix{0&1\cr-1&0\cr},B=\pmatrix{0&1\cr 0&0},\quad C=\pmatrix{1&0\cr 0&0\cr}.$$ The Lie algebras are all isomorphic. $A$ gives $SO(2)$, $B$ the additive group of reals, $C$ the multiplicative group of positive reals. – Jyrki Lahtonen May 13 '20 at 11:12
  • $B$ and $C$ happen to be isomorphic over reals, but if you switch to complex, you lose that. The main point is that the exponentiation of a Lie algebra is defined only on a particular representation of that Lie algebra. Different representations may give rise to non-isomorphic Lie groups. – Jyrki Lahtonen May 13 '20 at 11:17
  • If you assume that your Lie group is connected, then indeed every element of the Lie group can be written as the exponential of a vector in the Lie algebra. So the exponential map is surjective. This gives you from the information of the Lie algebra indeed the ability to tell what the product of any two elements of your Lie group is going to be. From this you might think that the group is fully determined by the Lie algebra, but it is not. The key to seeing this is that the exponential map does not need to be injective! As mentionned above the one-dimensional abelian Lie algebra can give both – Tychonoff3000 May 22 '20 at 22:23
  • the additive group of reals or the unit circle in the complex plane under multiplication. The former has injective exponential map while the latter does not. – Tychonoff3000 May 22 '20 at 22:25

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