Why isn't Likelihood a Probability Density Function?

The accepted answer is wrong. The likelihood is the function $\theta \mapsto L(\theta \mid x)=f(x \mid \theta)$ for a given $x$ in the observations space and $f(\cdot \mid \theta)$ is a Radon-Nikodym derivative of $P_\theta$ when the statistical model is given by a family of probabilities ${(P_\theta)}_{\theta\in\Theta}$ on the observations space.

In general there's not even a $\sigma$-field in the parameter space $\Theta$, hence the question "is the likelihood a pdf ?" is not even meaningful!

For more information see https://stats.stackexchange.com/questions/31238/what-is-the-reason-that-a-likelihood-function-is-not-a-pdf and https://stats.stackexchange.com/questions/29682/how-to-rigorously-define-the-likelihood

If $X$ is data and $m$ are the parameters, then the likelihood function $l(m) = p(X | m)$. I.e. it's $p(X | m)$, considered as a function of $m$.

Both $p(X|m)$ and $p(m|X)$ are pdfs: $p(X|m)$ is a density on $X$ and $p(m|X)$ is a density on $m$. But the likelihood is $p(X|m)$, not as a function of $X$ (it would indeed be a density as a function of X), but as a function of m. So it's not a pdf; in particular, it's not necessarily true that $$\sum_m p(X|m) = 1.$$

Edit: just to clarify, $p(m|X)$ isn't the likelihood. $p(X|m)$ is.

From a Bayesian perspective, the reason the likelihood function isn't a probability density is that you haven't multiplied by a prior yet. But once you multiply by a prior distribution, the product is (proportional to) the posterior probability density for the parameters.