Let $V$ be a finite-dimensional, real Banach space. We say $p:V\to\mathbb R$ is a polynomial if it is smooth with $m^{\mathrm{th}}$ derivative identically zero: $\nabla^m p\equiv 0$.
Let $G\hookrightarrow M_{n\times n}(\mathbb R)$ be a compact, real matrix group, embedded in some matrix algebra. We say that $p:G\to\mathbb R$ is a polynomial if $p$ is the restriction of some polynomial, $q:M_{n\times n}(\mathbb R)\to\mathbb R$, to $G$.
My question: is this definition of polynomials on $G$ independent of the embedding of $G$ into some matrix algebra? E.g., if we instead embed $G\hookrightarrow M_{(n+1)\times(n+1)}(\mathbb R)$, do we get the same space of 'polynomials on $G$'?
The question basically boils down to whether two continuous homomorphisms $G\to GL_n(\mathbb{R})$ can differ by something non-polynomial. it turns out that this is not the case.
First note that "polynomial in $x\in\mathbb{R}^n$" is the same as "polynomial in $z$ and $\overline{z}$ for $z\in\mathbb{C}^n$" so that I will directly look at the complex case.
I claim:
Since the set of those representations is only determined by $G$, this is no longer dependent on any particular embedding. Instead we now look at all continuous homomorphisms $\rho: G\to U(n)$ at the same time.
Now for the proof.
We start with classical representation theory: A compact group has an invariant Haar-probability-measure so that we can average many things over $G$. For example there is always a $G$-invariant scalar product $\langle,\rangle$ on $\mathbb{C}^n$ by averaging rotated copies of the euclidean scalar product $\langle,\rangle_2$: $$\langle x,y\rangle := \int_G \langle gx,gy\rangle_2 dg$$
By finding an orthonormal basis for this invariant scalar product, we find that all compact groups are conjugated to subgroups of $U(n)$.
The set of polynomial functions doesn't change under conjugation with any fixed matrix. That means: We can replace $GL_n(\mathbb{C})$ by $U(n)$ for our purposes.
What this gives us is Maschke's theorem: Now that we have unitary matrices, we can decompose $\mathbb{C}^n$ into a orthogonal, direct sum of $G$-invariant subspaces. Those continuous homomorphisms $\rho: G\to U(n)$ that cannot be decomposed in this way any further are called irreducible representations of $G$. We have just shown: Every continuous, finite-dimensional, complex representation of $G$ decomposes as a direct sum of irreducibles.
One can show more: These decompositions are actually unique up to conjugation.
Now observe that we can build now representations out of old ones: If $G\to Gl(V)$ is a representation on some complex vector space $V$, $G$ also acts on the dual space $V^\ast$, on direct sums $V^{\oplus n}$, tensor powers $V^{\otimes n}$ and any combination thereof.
One can also prove the following theorem: If $\psi: G\to Gl(V)$ is any fixed injective representation, then all irreducible representations occur as summands of somewhere in the induced representations $V^{\otimes n}\otimes (V^\ast)^{\otimes m}$.
This is important for our problem because polynomial functions $V\to\mathbb{C}$ are exactly the same as elements of the symmetric algebra $Sym(V^\ast) = \mathbb{C} \oplus V^\ast \oplus (V^\ast)^{\otimes 2} \oplus (V^\ast)^{\otimes 3}\oplus\cdots$ and polynomial functions in $\overline{z}$ are exactly the same as the symmetric algebra $Sym(V)$.
And that's basically it: You start with your embedding $\psi: G\to U_n(\mathbb{C})$. The dual representation is given by the $\psi^\ast: G\to U_n(\mathbb{C}), g\mapsto \overline{\psi(g)}$. Now you take all possible direct sums of all possible tensor products of $\psi$ and $\psi^\ast$. The coefficients of these matrices are all the possible polynomial functions evaluated at $\psi(g)$ and $\overline{\psi(g)}$. On the other hand these matrices are also conjugated to all possible representations of $G$ and since conjugation preserves the set of polynomial functions, this proves the theorem.