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I am trying to understand how components of a vector in the dual basis transform covariantly as mentioned in this quote.

If you seek to define a quantity (such as vector A) that remains invariant under a transformation of coordinates, you have a choice: you can combine superscripted (contravariant) components with subscripted (covariant) basis vectors, or you can combine subscripted (covariant) components with superscripted (contravariant) basis vectors.

Moreover I am trying to understand what transforms covariantly and contravariantly means exactly.

In the case of superscripted (contravariant) components with subscripted (covariant) basis vectors what I have inferred from the answers to my previous question is the following. Consider $V=\mathbb{R}^n$ and let $B$ denote the matrix of old basis vectors and $B'$ the matrix of new basis vectors. Then

$$B' = \begin{pmatrix}\vert & \vert & & \vert \\ b'_1 & b'_2 & \ldots & b'_n\\ \vert & \vert & & \vert \end{pmatrix} = BP = \begin{pmatrix}\vert & \vert & & \vert \\ b_1 & b_2 & \ldots & b_n\\ \vert & \vert & & \vert \end{pmatrix} P $$

and $[v]_{B'} = P^{-1}[v]_{B}$. Then contravariant means the use of the inverse of the matrix used for covariant. Hence the appearance of $P^{-1}$ and $P$. But this terminology of contravariant and covariant doesn't indicate whether the matrix $P^{-1}$ or $P$ multiples on the left or the right.

Now consider the the case of viewing the vector as (covariant) components with superscripted (contravariant) basis vectors. The matrix of dual basis vectors corresponding to $B$ is $D=(B^{-1})^T$. The matrix of dual basis vectors of $B'$ is $D' = ((B')^{-1})^T= ((BP)^{-1})^T=(B^{-1})^T(P^{-1})^T$. So that $D' = D(P^{-1})^T$ and $[v]_{D'} = P^T[v]_{D}$. Although the $P$ and $P^{-1}$ have switched places compared to the above as expected, I'm confused as there are transposes and yet the covariant contravariant terminology says nothing about this. Unless maybe I'm suppose to view the (covariant) components with superscripted (contravariant) basis vectors case as transposed so that all equations are transposed with everything written as row vectors so as to eliminate the transpose on $P$.

Surprisingly none of the books I've looked at talk about whether or when the matrix $P^{-1}$ or $P$ multiples on the left or the right or when to view things maybe as row vectors. Maybe I'm just looking at the wrong books?

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user782220
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1 Answers1

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Let us use upper indices to index row(-vector)s and lower index for column(-vector)s.

Let $x$ be an arbitrary vector, and $[x]$ its coordinates in the standard basis.

We will also consider two other bases $\{v_i\}$, $\{w_i\}$ and later the corresponding dual bases for the dual space of linear functionals (or 1-forms, or covectors).

Let $V=\left([v_1]\mid [v_2]\mid\ldots [v_n]\right)$ be the matrix made of coordinate columns-vectors of the first basis, and similarly $W$ for the second basis $\{w_i\}$.

If the coordinates of a vector in the two bases are related by $$ [x]_w=T[x]_v\tag{1}, $$ then, identifying the left-most and the right-most side in $V[x]_v=[x]=W[x]_w=W(T[x]_v)=(WT)[x]_v$, we have

$$V=WT\tag{2}$$

Let $V^{\prime}=\left(\begin{smallmatrix} [v^1]\\ [v^2] \\ \cdots\\ [v^n]\end{smallmatrix}\right)$ be the matrix of stacked row-vectors of the first dual basis, and similarly $W^{\prime}$ for the second dual basis $\{w^i\}$.

Using the dual basis we can write $[x]_v=\left(\begin{smallmatrix} v^1(x)\\ v^2(x) \\ \cdots\\ v^n(x)\end{smallmatrix}\right)=V^{\prime}[x]$ and similarly for the $[x]_w$.

From $TV^{\prime}[x]=T[x]_v=[x]_w=W^{\prime}[x]$ we have $$W^{\prime}=TV^{\prime}\tag{3}$$

Finally, for any dual vector (co-vector) $\alpha$ we can write $[\alpha]=[\alpha]_{w^{\prime}}W^{\prime}=[\alpha]_{w^{\prime}}TV^{\prime}$ and therefore, $$[\alpha]_{v^{\prime}}=[\alpha]_{w^{\prime}}T\tag{4}$$

The (1)-(4) tell us in which direction and from which side to apply the transformation matrix to the coordinates of vectors, covectors, and their bases.

rych
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