How does the representation of co-vectors change if we change the basis of a vector space $V$?

Question

I'm trying to understand how vectors, differential forms and multi-linear maps in general transform under change of coordinates. So I start with the simplest case of vectors. Here's my own attempt, please bear with me:

Suppose that $V$ is a vector space and $\alpha=\{v_1,\cdots,v_n\}$ and $\beta = \{w_1,\cdots,w_n\}$ are two ordered bases for $V$. $\alpha$ and $\beta$ give rise to the dual bases $\alpha^*=\{v^1,\cdots,v^n\}$ and $\beta^*=\{w^1,\cdots,w^n\}$ for $V^*$ respectively.

If $[T]_{\beta}^{\alpha}=[\lambda_{i}^{j}]$ is the matrix representation of coordinate transformation from $\alpha$ to $\beta$, i.e.

$$\begin{bmatrix} w_1 \\ \vdots \\ w_n \end{bmatrix}= \begin{bmatrix} \lambda_1^1 & \lambda_1^2 & \dots &\lambda_1^n \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_n^1 & \lambda_n^2 & \cdots & \lambda_n^n\end{bmatrix} \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix}$$

What is the matrix of coordinate transformation from $\alpha^*$ to $\beta^*$?

We can write $w^j \in \beta^*$ as a linear combination of basis elements in $\alpha^*$:

$$w^j=\mu_{1}^{j}v^1+\cdots+\mu_n^{j}v^n$$

We get a matrix representation $[S]_{\beta^*}^{\alpha^*}=[\mu_{i}^{j}]$ as the following:

$$\begin{bmatrix} w^1 & \cdots & w^n \end{bmatrix}= \begin{bmatrix} v^1 & \cdots & v^n \end{bmatrix}\begin{bmatrix} \mu_1^1 & \mu_1^2 & \dots &\mu_1^n \\ \vdots & \vdots & \ddots & \vdots \\ \mu_n^1 & \mu_n^2 & \cdots & \mu_n^n\end{bmatrix} $$

We know that $w_i = \lambda_{i}^1v_1+\cdots+\lambda_{i}^nv_n$. Evaluating this functional at $w_i \in V$ we get:

$$w^j(w_i)=\mu_{1}^{j}v^1(w_i)+\cdots+\mu_n^{j}v^n(w_i)=\delta_{i}^j$$ $$w^j(w_i)=\mu_{1}^{j}v^1(\lambda_{i}^1v_1+\cdots+\lambda_{i}^nv_n)+\cdots+\mu_n^{j}v^n(\lambda_{i}^1v_1+\cdots+\lambda_{i}^nv_n)=\delta_{i}^j$$ $$w^j(w_i)=\mu_{1}^{j}\lambda_{i}^1+\cdots+\mu_n^{j}\lambda_{i}^n=\sum_{k=1}^n\mu_{k}^j \lambda_{i}^k=\delta_{i}^j$$

But $\sum_{k=1}^n\mu_{k}^j \lambda_{i}^k$ is the $(i,j)$ entry of the matrix product $TS$. Therefore $TS=I_n$ and $S=T^{-1}$.

If we want to write down the transformation from $\alpha^*$ to $\beta^*$ as column vectors instead of row vector and name the new matrix that represents this transformation as $U$, we observe that $U=S^{t}$ and therefore $U=(T^{-1})^t$.

Therefore if $T$ represents the transformation from $\alpha$ to $\beta$ by the equation $\mathbf{w}=T\mathbf{v}$, then $\mathbf{w^*}=U\mathbf{v^*}$.

The important case is when $T=(T^{-1})^t$ which happens if and only if $T$ is an orthonormal transformation.

What I don't understand is why the transformation from $\alpha$ to $\beta$ is called a contravariant transformation while the transformation from $\alpha^*$ to $\beta^*$ is called a covariant transformation. Would you please elaborate on this important point? It's been driving me crazy for the last two days.

Thanks in advance.

Answer 1

If we have some vector $u$, then its components $(u^j)^n_{j=1}$ in the basis $\alpha$ can be found by applying the dual basis $\alpha^*$ to $u$. $$u^j=v^j(u)$$

So when we change from $\alpha$ to $\beta$ the components change by $$u^j=v^j(u)\longmapsto w^j(u)=(S\mathbf{v})^j(u)=\sum_i{\mu^j}_iu^i$$ So the components of vectors change by left multiplication by the matrix $S$ (with components ${\mu^j}_i$).

Similarly, covectors $f$ have components $f_i$ which transform by $$f_i\longmapsto \sum_j{\lambda_i}^jf_j$$ So the components of covectors change by left multiplication by the matrix $T$ (with components ${\lambda_i}^j$).

Components that transform according to $S$ are thought of are changing in the same way as the components for vectors, hence they are co-variant ("varying with").

Components that transform according to $S^{-1}=T$ are transforming in the opposite way and are therefore contra-variant ("varying against").

Answer 2

Let us use upper indices for row(-vectors) and lower index for column(-vectors).

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

Let $V=\left([v_1]\mid [v_2]\mid\ldots [v_n]\right)$ be the matrix of stacked 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 $$[x]_w=T[x]_v,$$ then, identifying left-most and right-most sides in $V[x]_v=[x]=W[x]_w=W(T[x]_v)=(WT)[x]_v$ we have $$V=WT$$

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 $$W^{\prime}=TV^{\prime}$$

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