# Dual to dual

Just realized how double dual isn't really the original space even in simple Euclidean case (which I was aware of, but some how didn't feel I was understanding): with vector being a column $x\in\mathbb{R}^n$, dual a row $x^\top\in(\mathbb{R}^n)^*$, the double dual $x^{\top\top}$ is indeed a column, except when it acts on the rows in multiplication it goes TO THE RIGHT and not to the left:

\begin{equation*} \begin{split} x^\top(y) &= x^\top y,\\ x^{\top\top}(y^{\top}) &= y^\top x^{\top\top} = y^\top x. \end{split} \end{equation*}

This contrasts with rows acting on columns, where the dual (the row) acts from the left.

So exciting.

Update! It's much more than that! If we treat $\mathbb{R}^n$ as a manifold, it turns out then that its tangent space looks more like double dual $(\mathbb{R}^n)^{**}$ rather than $\mathbb{R}^n$ or $(\mathbb{R}^n)^*$, because when we consider a tangent vector acting on scalar functions $\mathbb{R}^n\to\mathbb{R}$ in the special -- linear -- case, the tangent goes on the right and it does so as a column. Take a scalar function $(\mathbb{R}^n)^* \ni a^\top : \mathbb{R}^n \to \mathbb{R}$. Then a tangent $X\in\mathcal{T}\mathbb{R}^n$ should act on $a$ from the right:

\begin{equation*} X(a) = \left.\partial_t(t\mapsto a(x) + a^\top t X)\right|_{t=0} = a^\top X. \end{equation*}

Here $x\in\mathbb{R}^n$ denotes $X\in\mathcal{T}\mathbb{R}^n$ casted to $\mathbb{R}^n$