For a fixed dimension $n$ and degree $k$, let $H$ be the space of all real homogeneous harmonic polynomials of degree $k$ in $n$ variables. We equip $H$ with the inner product $\langle f,g\rangle = \int_{S^{n-1}} f(x) g(x) d\sigma(x)$, where $\sigma$ is the surface measure on the unit sphere $S^{n-1}$. Let $\{v_i\}_i$ be an orthonormal basis for $H$.
By rotation invariance, we can show that the function $S^{n-1} \times S^{n-1} \to \mathbb R$ defined by $(x,y) \mapsto \sum_i v_i(x) v_i(y)$ only depends on $\langle x,y\rangle$. Thus, there exists a function $Q : [-1,1] \to \mathbb R$ such that
$$Q(\langle x,y\rangle) = \sum_i v_i(x) v_i(y).$$
More concretely, we have
$$Q(t) = \sum_i v_i(1,0,\ldots, 0) v_i(t, \sqrt{1-t^2},0,\ldots,0). \tag{$*$}$$
From $(*)$ and the fact that $v_i \in H$, we can see that $Q(t)$ is a polynomial in $t$ and $\sqrt{1-t^2}$.
My question is: From $(*)$, can we see that $Q(t)$ is a polynomial in $t$ of degree (exactly) $k$? (In other words, I am wondering if we can use $(*)$ as the definition of Gegenbauer polynomials.)
Here is a partial answer. Define
$$P(t,u) = \sum_i v_i(1,0,\ldots, 0) v_i(t, u,0,\ldots,0).$$
This is a homogeneous polynomial of degree $k$ in two variables. Furthermore, by rotation invariance, $P(t,u) = P(t,-u)$ whenever $t^2 + u^2 = 1$. By homogeneity, $P(t,u) = P(t,-u)$ for all $(t, u) \in \mathbb R^2$. This implies that all the powers of $u$ that appear in the polynomial $P(t,u)$ are even. Thus, $Q(t) = P(t, \sqrt{1-t^2})$ is a polynomial in $t$ of degree at most $k$.
(I'm not sure how to show the degree of $Q$ is exactly $k$.)