# Finite representability of $\ell_1$ and $c_0$

It follows from Krivine's theorem for a given Banach space $X$, either some $\ell_p$ or $c_0$ is finitely represented in $X$. Since finite dimensional $\ell_1$ and $\ell_\infty$-spaces are dual to each other, can we expect that

$\ell_1$ is finitely representable in $X$ if and only if $c_0$ is finitely representable in $X^*$?

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No, $X=c_0$ with $X^*=\ell_1$ is a counterexample. The space $\ell_1$ isometrically embeds into $\ell_\infty$. For any Banach space $Y$ the bidual $Y^{**}$ is finitely representable in $Y$ (the principle of local reflexivity). Since $\ell_\infty=c_0^{**}$ we conclude that $\ell_\infty$, and a fortiori $\ell_1$ is finitely representable in $X$.
On the other hand, $c_0$ is not finitely representable in $X^*$. Indeed, $\ell_1$ has cotype $2$, while $c_0$ does not; in fact both type and cotype of $c_0$ are trivial because $\ell_\infty$ is finitely representable in it.
But the other direction holds: if $\ell_1$ is not finitely representable in $X$, then the type of $X$ is strictly greater than $1$, by the Maurey-Pisier theorem. It follows that the cotype of $X^*$ is finite, and therefore $c_0$ is not finitely representable in $X^*$.
Added a reflexive example. Let $X=\bigoplus_{n=2}^\infty \ell_n^n$; so that $X^*=\bigoplus_{n=2}^\infty \ell_{n/(n-1)}^n$. (The direct sums are $\ell_2$-normed.) Again, $c_0$ is not finitely representable in $X^*$ because $X^*$ has cotype $2$. As for $X$, it has arbitrarily large subspaces that are arbitrarily close to being in $\ell_\infty$, which implies that $\ell_1$ is finitely representable in it.