Let $X_1, \ldots, X_d$ be $d$ independent Gaussian $N(0,1)$ random variables, and let $$Y=\frac{1}{\| X \|} (X_1, \ldots ,X_d)$$ Clearly $Y$ lies on the surface of the sphere $S^{d-1}$.
Let the vector $Z \in \mathbb{R}^k$ be the projection of $Y$ onto its first $k$ coordinates, and let $L = \| Z \|^2$.
Clearly the expected squared length of $Z$ is $\mu = E[L] = \dfrac{k}{d}$.
Question:
(1) Why $E[L] = \dfrac{k}{d}$?
(2) Does any of the information above imply $X_1^2 + X_2^2 + \cdots + X_d^2=1$?
Remark: The above paragraph is obtained from this paper.
PS: For the moment I've struck out what I wrote below, however some trivial modifications should fix it. You wrote ${}\in\mathbb R^k$. When I see the word "projection" I normally think of what I wrote below, and I believe that's how the word is most frequently used in linear algebra and geometry.
Here is the projection: $$ (Y_1,\ldots,Y_k,Y_{k+1},\ldots, Y_d) \mapsto (Y_1,\ldots,Y_k,0,\ldots, 0) = Z. $$ Then we have $$ Z^2 = Y_1^2+\cdots+Y_k^2 + \underbrace{0^2 + \cdots + 0^2}_\text{all 0s}. $$ $$ \operatorname{E}(\|Z\|^2) = \operatorname{E}(Y_1^2)+\cdots+\operatorname{E}(Y_k^2) = k\operatorname{E}(Y_1^2). \tag 1 $$ And $$ d\operatorname{E}(Y_1^2) = \operatorname{E}(Y_1^2)+\cdots+\operatorname{E}(Y_d^2) = \operatorname{E}(Y_1^2+\cdots+Y_d^2) = \operatorname{E}(1) = 1 $$ So $\operatorname{E}(Y_1^2) = \dfrac 1 d$, and so by $(1)$, $$ \operatorname{E}(\|Z\|^2) = \frac k d. $$