# Why linear interpolation not appropriate for volatility surface construction?

We know linear interpolation is not appropriate for constructing a surface, but why?

In the book, "Foreign Exchange Option Pricing: A Practitioners Guide", the author writes:

native linear interpolation with regard to time can lead to unrealistic forward volatility dynamics... this implies a negative forward variance between ...

I am not sure I understand the reasoning. Why does linear interpolation imply negative forward volatility ? Can anyone provide a better explanation? Is there any other reason that the simple linear interpolation should not be used?

Note that total implied variance defined as $$V(T,K) = T\Sigma(T,K)^2$$ should be an increasing function of $T$. Otherwise you have a calendar arbitrage (sell the call with shorter expiry and buy the cheap longer one).
If you interpolate linearly your implied volatility is $$\Sigma(T,K) = w\Sigma(T_i,K) + (1-w)\Sigma(T_{i+1},K)$$ with weight $w = \frac{T_{i+1}-T}{T_{i+1}-T_i}$. This can also be written as $$\Sigma(T,K) = \Sigma(T_i,K) + s(T-T_i)$$ with slope $s = (\Sigma(T_{i+1},K)-\Sigma(T_{i},K))/(T_{i+1}-T_i)$. Note that $s$ can be negative, i.e. $\Sigma(T_{i+1},K) < \Sigma(T_{i},K)$ even in an arbitrage-free situation: $V(T_{i+1},K) \ge V(T_{i},K)$.
Now all you have to do is check for calendar arbitrage: $$\partial_T V(T,K) \geq 0$$ A simple computation will show you that the lhs is a 2nd order polynomial in $T$ and that it can turn negative.