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Let $S^{d-1}$ be the unit sphere in $\mathbb{R}^d$ and $\operatorname{Pol}_{\leq n}(S^{d-1})$ be the space of polynomial functions of degree at most $n$. I'm trying to understand the irreducible decomposition of $\operatorname{Pol}_{\leq n}(S^{d-1})$ under the action of the orthogonal group $O(d)$. My main reference is the book "Approximation Theory and Harmonic Analysis on Spheres and Balls" by Dai and Xu.

Let $\mathcal{P}_n^d$ be the space of homogeneous polynomials of degree $n$ in $d$ variables. Let $\operatorname{Harm}_n^d$ be the subspace of $\mathcal{P}_n^d$ consisting of all homogeneous harmonic polynomials of degree $n$, i.e., $$ \operatorname{Harm}_n^d=\{f\in\mathbb{R}[x_1,\dots,x_d]\mid f\text{ homogeneous}, \operatorname{deg}(f)=n, \Delta f=0\}. $$

I understand how we can prove by induction that the space $\mathcal{P}_n^d$ decomposes as $$ \mathcal{P}_n^d=\bigoplus_{j=0}^{\lfloor n/2\rfloor} ||x||^2 \operatorname{Harm}_{n-2j}^d.\tag{1} $$ In Dai, Xu, it is written that because of this decomposition, we obtain the following decompositions of the space $\operatorname{Pol}_{\leq n}(S^{d-1})$ $$ \operatorname{Pol}_{\leq n}(S^{d-1})=\bigoplus_{k=0}^n \left.\operatorname{Harm}_k^d\right|_{S^{d-1}}\tag{2} $$ and $$ \operatorname{Pol}_{\leq n}(S^{d-1})=\left.\mathcal{P}_n^d\right|_{S^{d-1}}\oplus \left.\mathcal{P}_{n-1}^d\right|_{S^{d-1}}.\tag{3} $$ I don't see how we get the decompositions (2) and (3) by using (1).

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    $\begingroup$ I think it is a consequence of $\|x\|=1$ on the sphere. $\endgroup$ – Giuseppe Negro Sep 19 '18 at 10:16
  • $\begingroup$ You're right that is the reason. Thanks! $\endgroup$ – user160919 Sep 19 '18 at 10:43
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I forgot that we look at the restriction of the polynomials to the unit sphere. Therefore, we have $||x||=1$. A polynomial in $\operatorname{Pol}_{\leq n}(S^{d-1})$ is of degree at most $n$ and hence, it can be written as a linear combination of homogeneous polynomials of degree k, where $k=0,1,\dots,n$. Therefore, we have the decomposition $$ \operatorname{Pol}_{\leq n}(S^{d-1}) =\bigoplus_{k=0}^n\left.\mathcal{P}_k^d\right|_{S^{d-1}}. $$ By using the decomposition (1) and $||x||=1$, we obtain $$ \operatorname{Pol}_{\leq n}(S^{d-1}) =\bigoplus_{k=0}^n \bigoplus_{j=0}^{\lfloor k/2\rfloor}\left.\operatorname{Harm}_{k-2j}^d\right|_{S^{d-1}}=\bigoplus_{k=0}^n \left.\operatorname{Harm}_k^d\right|_{S^{d-1}}. $$ This immediately gives us $$ \operatorname{Pol}_{\leq n}(S^{d-1})=\left.\mathcal{P}_n^d\right|_{S^{d-1}}\oplus \left.\mathcal{P}_{n-1}^d\right|_{S^{d-1}}. $$

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