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My question is easily stated: are all continuous finite-dimensional real representations of $\text{SL}_n(\mathbb{R})$ algebraic representations?

This is false if you drop the word "continuous" (e.g. using noncontinuous field automorphisms of $\mathbb{R}$).

It is also false if you work with $\text{GL}_n(\mathbb{R})$ instead of $\text{SL}_n(\mathbb{R})$. For instance, there are many examples in the answers to this question. However, all of the continuous examples there use the determinant in some way, so they don't give examples of non-algebraic representations of $\text{SL}_n(\mathbb{R})$.

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    $\begingroup$ Every field automorphism of $\mathbf{R}$ is continuous (however you get representation using non-continuous field embeddings into $\mathbf{C}$). $\endgroup$
    – YCor
    Jan 22, 2020 at 6:13
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    $\begingroup$ The answer to your question is yes, and more generally true for every connected semisimple Lie group, and even perfect is enough. This is due to the fact that every perfect subalgebra of matrices is the Lie algebra of an algebraic subgroup. This can be applied to the (Lie algebra of the) graph of your representation. $\endgroup$
    – YCor
    Jan 22, 2020 at 6:19
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    $\begingroup$ @YCor: Maybe one should mention that this graph is a closed subgroup of a Lie group, hence a Lie subgroup, and therefore algebraic by your argument. $\endgroup$
    – abx
    Jan 22, 2020 at 9:40
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    $\begingroup$ @YCor: "Every field automorphism of R is continuous": Even more is true: The identity is the only field automorphism of $\mathbb{R}$. $\endgroup$
    – Todd Leason
    Jan 22, 2020 at 9:49
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    $\begingroup$ @YCor, the implication "perfect matrix Lie algebra $\implies$ Lie algebra of algebraic group" doesn't seem to be easily googlable. Could convert your comment into a short answer with a reference? $\endgroup$
    – Igor Khavkine
    Jan 22, 2020 at 11:05

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