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I know the definitions of algebraic and Lie groups. I know the difference between them in terms of definition; loosely, the first is a variety plus group, while the second is a smooth manifold plus group.

However, in terms of examples, the main examples are almost the same. Also, the tangent space of both of them at the identity is a Lie algebra. In addition, both are studied in the big title "Lie Theory" and both are classified by Dynkin diagrams, etc.

I know there might be some examples of Lie groups that are not algebraic and vice-versa. However, the theory behind them looks to be almost the same. So my question, and I hope it is clear, what is the real difference between them? Do they coincide in some cases and differ in some?

F.H.A
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    One of the biggest differences is that for an algebraic group, you can often plug in an underlying field which is not $\mathbb{R}$ or $\mathbb{C}$, for example an algebraically closed field of characteristic $p > 0$. This leads to all sorts of interesting modular representation theory. – Joppy Feb 24 '22 at 08:07
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    Both concepts have a huge overlap and are mostly motivated by the same examples but there are examples in both direction which don't have a counterpart on the other side. On the lie group side there are groups which are in a sense "exponential" and therefore not algebraic: see https://math.berkeley.edu/~reb/courses/261/3.pdf for examples. And on the other side as @Joppy pointed out you can generalize algebraic groups to arbitrary fields, so they don't have a notion of smoothness anymore. – Maik Pickl Feb 24 '22 at 08:19
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    While I think that the above two comments are helpful, and good, I would like to point out two technicalities: a) 'plugging' both $\mathbb{C}$ and $\mathbb{F}_p$ is something that you can do (if for instance you group is *defined over $\mathbb{Z}$) but not something that makes sense in general -- e.g. if $G$ is a unitary group for a ramified extension of $p$-adic local fields), b) algebraic groups do have a notion of smoothness and, in fact, every group in characteristic 0 is smooth by a.theorem of Cartier. Of course, this is smoothness in an algebro-geometric sense. – Alex Youcis Feb 24 '22 at 08:47
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    I'd also like to mention a case when they agree: compact connected Lie groups over $\mathbb{R}$ are the same thing as connected reductive anisotropic groups over $\mathbb{R}$ -- the former is a Lie group thing (obviously) and the latter is an object of algebraic group theory. This is significant as most people's first introduction to the theory of Lie groups is via the compact theory -- so if that's been your experience, it's not surprising all your Lie groups look algebraic. There's also a correspondence between anisotropic groups over $\mathbb{R}$ and reductive groups over $\mathbb{C}$ – Alex Youcis Feb 24 '22 at 08:49
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    --every complex reductive (algebraic) group over $\mathbb{C}$ has a unique anisotropic form over $\mathbb{R}$. It's for this reason that the theory of complex reductive groups and compact Lie groups looks so similar. The above also explains why to find an example of a Lie group which is not algebraizable you look for examples over $\mathbb{R}$ which are not compact (the most well-known being the universal cover of $\mathrm{SL}_2(\mathbb{R})$). All of this is covered very thoroughly in Appendix D of this: http://math.stanford.edu/~conrad/papers/luminysga3.pdf – Alex Youcis Feb 24 '22 at 08:51
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    but I fear that this might be too advanced a reference if you are not faimilar with algebraic geometry. – Alex Youcis Feb 24 '22 at 08:53
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    On an elementary level, if you proceed to solvable Lie groups (often motivated by geometry) you will see that many are not algebraic - see here for posts on non-algebraic Lie algebras, or on MO, e.g. here. – Dietrich Burde Feb 24 '22 at 09:17
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    "I know there might be some examples of Lie groups that are not algebraic". Actually, there really are such examples. They have been dicussed here in various posts, see for example here. Also, the $5$-dimensional solvable Lie group from the MO post (Bourbaki) is such an example. – Dietrich Burde Feb 24 '22 at 09:52

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