Does AIC criterion for variable selection in the least square method require the data to be normally distributed? I'm studying Linear Regression from the book A Modern Approach to Regression with R by Simon J. Sheather (the 2009 edition). The chapter 7 (page 228), in which different criterion for variable selection is introduced, the author said:

The other three criteria are based on likelihood theory when both the predictors and the response are normally distributed.

Here the three criteria in the mention are $AIC, AIC_C, BIC$. Then by assuming that $y, x_{1i},\dots,x_{pi}$ each follows a normal random variable, he reasons that $y|x_{1i},\dots x_{pi}\sim N(\beta_0+\dots+\beta_px_{pi},\sigma^2)$. From this point, he proceeds to find the Maximum Likelihood $L(\beta_0,\dots,\beta_p,\sigma|Y)$.
I know from this thread that it isn't AIC which demands the assumption of normally distributed residuals but the Least Square Regression method. But what about the assumption of normally distributed observations? Is it required for the criterion to work properly?
After the introduction, the author provides some examples of the application of the criteria. I had a look at the Bridge Construction data (page 234, the dataset is available online) in which the predictor $\log(Spans)$ of the model fails the Shapiro-Wilk test, which confuses me much further.
It would be appreciated if somebody could help me clear this confusion.
 A: Ok, there are a bunch of issues here, including the difference between necessary and sufficient conditions.
Least-squares regression does not require a Normal distribution; it requires that you want to minimise squared error.  However, least-squares regression is maximum likelihood when the errors are Normal, and without Normal errors it is usually not maximum likelihood[0]
AIC, the thing derived by Akaike, is computed from the loglikelihood (or from the deviance).  Computing it from the residual sum of squares, as least-squares regression software does, is correct only when the errors are Normal and the deviance is the $\log\mathrm{RSS}$, up to constants.
If you are using least-squares regression and the errors aren't Normal, the number that your software spits out with the label AIC is not, strictly speaking, Akaike's Information Criterion. It does, however, perform roughly the same role.
The analog of AIC for misspecified or 'working' models (what you get if you cross AIC with the sandwich (aka robust, aka HC) variance estimator) is Takeuchi's Information Criterion.  It's a penalised 'working' loglikelihood -- that is, it's a penalised version of the number your software spits out with the label 'logL' -- but the penalty is not $2p$.
The penalty is twice the trace (sum of diagonal elements) of a matrix. The matrix is $I^{-1}J$ where $I$ is minus the expectation of the second derivative of the 'working' loglikelihood and J is the variance of the first derivative of the 'working' loglikelihood.  If the 'working' loglikelihood is a genuine, honest-to-Fisher loglikelihood, plus some regularity conditions, the information identity (second Bartlett identity) says that $I=J$. That means $I^{-1}J$ is an identity matrix and its trace is the number of variables in the model. We recover a penalty of $2p$.
If the 'working' loglikelihood is something like the residual sum of squares in non-Normal data, the penalty does not reduce exactly to $2p$. However, $I^{-1}J$ can also be written as $V_m^{-1}V_s$ where $V_m$ is the model-based variance estimator and $V_s$ is the sandwich (aka robust, aka HC) variance estimator. If you don't have Normal errors but do have constant error variance (homoskedasticity) these will be about the same, and the correct penalty will be close to $2p$.  This indicates, as is often the case, that non-Normality is not of one of the top things to worry about.
Even with moderate amounts of heteroskedasticity the penalty typically won't be far off $2p$.  You'd probably be better off fitting a model that described the data better and using its AIC; you'd also be better off using the Takeuchi AIC. But the (admittedly, wrong) AIC shouldn't be grossly misleading unless you've got lots of heteroskedasticity.
If you're interested in reading more on this, Alastair Scott and I wrote a paper on extending the AIC to survey sampling, which is like the Takeuchi extension only more so.
[0]: usually not: consider a $2\times 2$ table of binary $X$ by binary $Y$. The obvious estimator of the mean of $Y$ for each value of $X$ is both the least-squares estimator (it's the mean) and the maximum likelihood estimator.
