Computation of variance in a feature space. I am trying to read this paper on video summarization, and I am running into difficulties understanding some of notation.
Given a sequence of vectors $x_t$, a kernel function is defined with the feature space  $\mathcal{H}$ and variance between $x_t$ and $x_{t+1}$ is computed as 
$v_{t_{i},t_{i+1}} = \sum_{t=t_{i}}^{t_{i+1}-1}||\phi(x_{t})-\mu_{i}||^2_{\mathcal{H}}$
Later, in the algorithm section, the above variance is calculated as something that implies
$v_{t,t+d} = \sum_{i=t}^{t+d-1}K(x_i,x_i) - \frac{1}{d}\cdot\sum_{i,j=t}^{t+d-1}K(x_i,x_j)$
I am having trouble seeing the connection, though I do know that $K(x,y)=\phi(x)^T\phi(y)$
 A: As you've noted, the authors switched their notation. If you define $t:=t_i$,  $d:=t_{i+1}-t_i$, and (for brevity) $w_k:=\phi(x_{t_i+k-1})$ for $k=1,\ldots,d$, then the formula
$$\mu_i=\frac{\sum_{t=t_i}^{t_{i+1}-1}\phi(x_t)}{t_{i+1}-t_i}\tag1
$$becomes
$$\mu_i=\frac{\sum_{k=1}^d w_k}{d}\tag2
$$ (so $\mu_i$ is the mean of the $w$'s), while the formula
$$
v_{t_i,t_{i+1}} = \sum_{t=t_i}^{t_{i+1}-1}\|\phi(x_t)-\mu_i\|^2\tag3
$$
becomes
$$
v_{t,t+d}=\sum_{k=1}^d\|w_k-\mu_i\|^2,\tag4
$$
i.e. (4) is the variance of the $w$'s. You can expand (4) using $\|a\|^2=a^Ta$ to find
$$\|w_k-\mu_i\|^2=w_k^Tw_k-w_k^T\mu_i-\mu_i^Tw_k+\mu_i^T\mu_i,\tag5$$
then sum (5) from $k=1$ to $d$ to obtain [using (2)]
$$
\sum_{k=1}^d\|w_k-\mu_i\|^2=\sum_{k=1}^dw_k^T w_k -\frac1d\left(\sum_{k=1}^d w_k\right)^T\left(\sum_{k=1}^d w_k\right),\tag6
$$
which is the analog of the familiar formula for the variance of a univariate list. Now undo all the new notation, remembering that $K(x,y)=\phi(x)^T\phi(y)$, and you should arrive at that final formula.
