Editing Linear algebraic group (section)

===Lie groups===
For a linear algebraic group ''G'' over the real numbers '''R''', the group of real points ''G''('''R''') is a [[Lie group]], essentially because real polynomials, which describe the multiplication on ''G'', are [[smooth function]]s. Likewise, for a linear algebraic group ''G'' over '''C''', ''G''('''C''') is a [[complex Lie group]]. Much of the theory of algebraic groups was developed by analogy with Lie groups.

There are several reasons why a Lie group may not have the structure of a linear algebraic group over '''R'''.

*A Lie group with an infinite group of components G/G<sup>o</sup> cannot be realized as a linear algebraic group.
*An algebraic group ''G'' over '''R''' may be connected as an algebraic group while the Lie group ''G''('''R''') is not connected, and likewise for [[simply connected]] groups. For example, the algebraic group ''SL''(2) is simply connected over any field, whereas the Lie group ''SL''(2,'''R''') has [[fundamental group]] isomorphic to the integers '''Z'''. The double cover ''H'' of ''SL''(2,'''R'''), known as the '''[[metaplectic group]]''', is a Lie group that cannot be viewed as a linear algebraic group over '''R'''. More strongly, ''H'' has no faithful finite-dimensional representation. 
*[[Anatoly Maltsev]] showed that every simply connected nilpotent Lie group can be viewed as a unipotent algebraic group ''G'' over '''R''' in a unique way.<ref>Milne (2017), Theorem 14.37.</ref> (As a variety, ''G'' is isomorphic to [[affine space]] of some dimension over '''R'''.) By contrast, there are simply connected solvable Lie groups that cannot be viewed as real algebraic groups. For example, the [[universal cover]] ''H'' of the semidirect product ''S''<sup>1</sup> ⋉ '''R'''<sup>2</sup> has center isomorphic to '''Z''', which is not a linear algebraic group, and so ''H'' cannot be viewed as a linear algebraic group over '''R'''.