Weights in Adaboost The Adaboost algorithm is shown below.
I have a couple of questions.
First, I am a little confused on how the weights for each training sample is applied when using Adaboost. From looking at the algorithm it appears that the each sample's weight is only used when computing the error. Each input $x_i$ and true value $y_i$ are not altered?
Second, it seems the weights $w_i$ are not guaranteed to sum to 1? I was under the impression that they do, but in part (b) of the algorithm below, the summation in the numerator makes me think it doesn't always sum to 1 (otherwise they wouldn't have explicitly written it in the denominator?). Further inspection of part (d) also supports this. Is there an significance to not allowing the sample weights to sum to 1 vs. having a formulation that's normalized?
Do the weights $\alpha_i$ sum to 1?

Taken from Elements of Statistical Learning
 A: As for your first question, the only purpose of the weights is to make misclassified points in this iteration contribute more to the loss in the next iteration so the weak classifier can learn from its mistakes (now that they're emphasized with the weights). You'd never need to alter the individual $x_i$'s and $y_i$'s.
But typically, the weak classifier may not be using misclassification loss as the algorithm assumes. That is, $G_m(x)$ may not be necessarily minimizing
$$err_m=\frac{\sum_{i=1}^N w_iI(y_i\neq G_m(x_i))}{\sum_{i=1}^N w^i}$$
and instead it might be minimizing some other loss function. So in that case a similar effect can be achieved if we alter the dataset itself such that the presence (repetitions) of points with higher weights is more than those with lower weight and this is typically done by rejection sampling in which case it makes sense to think about the sample weights as a distribution.
For your second question, no. In the algorithm you provided they're only guaranteed to sum to 1 in the first iteration. However, you can always divide all the weights by some number so that they sum to 1 and it acts as a distribution. Notice how $err_m$ does not change when you multiply all the weights by some scalar. Thus, this step should have no significance unless you need it for something like rejection sampling.
Finally, no. The $\alpha$'s do no not essentially sum to 1. In fact each individual $\alpha$'s value can exceed 1 if the loss due to its corresponding classifier is small enough. If you were using a different voting scheme (e.g. by removing the $sgn$) then you might be interested in normalizing the $\alpha$'s by the end of the algorithm so that if every single weak classifier voted for 1 then you'd get 1 (same for -1) otherwise it will be a value between -1 and 1 and if see something like 0.1 you know that this one was almost a tie so besides of reporting '1' as $sgn$ would do you can also report that "well, the classifier is not really sure about this one but the guess is that its 1".
