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 \(\mathbb{R}^n\)

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