Clustering of standard errors in Fixed Effects models

Why is there still a block structure in the covariance-variance matrix, and consequently a need for clustering of standard errors in fixed-effects models? Shouldn't demeaning solve the serial correlation issue? Derivations and intuitive explanations are both appreciated!

• If serial correlation in the error term ($u_i + v_{it}$) solely comes from the time invariant part ($u_i$), you are right. But serial correlation may exist in $v_{it}$ (the time-varying part), which is not dealt with by the within-group transformation. Oct 8 '21 at 23:40

Consider the following specification: $$Y_{i,g} = X_{i,g}\beta + u_{i,g}$$ Where the residuals have different mean across groups and have within group correlation: \begin{align*} &\mathbb{E}(u_{i,g}) = \alpha_g,\\ &cov(u_{i,g} u_{j,g}) = \rho_{i,j},\\ &cov(u_{i,g}, u_{j,g'}) = 0 \text{ for } g \ne g' \end{align*} Taking means gives: $$\mathbb{E}(Y_{i,g}) = \mathbb{E}(X_{i,g}) \beta + \alpha_g$$ Now use notation $$\widehat{Z} = Z - \mathbb{E}(Z)$$ for the de-meaned variable. Then: $$\widehat{Y_{i,g}} = \widehat{X_{i,g}} \beta + \widehat{u_{i,g}}$$ Now the mean of $$\widehat{u_{i,g}}$$ has become zero but there is still correlation within each group: \begin{align*} &cov(u_{i,g}, u_{j,g}) = \mathbb{E}(\widehat{u_{i,g}},\,\ \widehat{u_{j,g}}) = \rho_{i,j},\\ &cov(u_{i,g}, u_{j,g'}) = \mathbb{E}(\widehat{u_{i,g}},\,\ \widehat{u_{j,g'}}) = 0 \text{ for } g \ne g' \end{align*} So de-meaning does not get rid of the within-group correlation.