1. Dual basis in $\mathbb R^2$
Enter basis vectors $v_1,v_2$. If $P=[v_1\ v_2]$, then the rows of $P^{-1}$ are the dual basis functionals.
Basis matrix $P=[v_1\ v_2]$
2. Coordinate extraction by measurements
Enter a vector $x$. The dual basis returns the coordinate vector $[x]_{\mathcal B}=P^{-1}x$.
3. Annihilator in $\mathbb R^3$
Enter two spanning vectors for a subspace $U\subseteq\mathbb R^3$. The annihilator $U^\circ$ consists of row vectors $\ell$ satisfying $\ell u_1=\ell u_2=0$.
Columns $u_1,u_2$
4. Dual map and transpose
If $T(x)=Ax$, then the dual map $T^*$ is represented by $A^T$.
This is the same transpose operation that appears in coboundary matrices.
5. Cohomology of a triangle
Compare a hollow triangle with a filled triangle.
6. Boundary and coboundary matrices
7. Independent-study practice tasks with answers
Task A. Find the dual basis for $v_1=(1,1)$ and $v_2=(1,2)$.
Answer: $P^{-1}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$, so $\varphi^1(x,y)=2x-y$ and $\varphi^2(x,y)=-x+y$.
Answer: $P^{-1}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$, so $\varphi^1(x,y)=2x-y$ and $\varphi^2(x,y)=-x+y$.
Task B. Find one functional that vanishes on $U=\operatorname{span}\{(1,1,0),(0,1,1)\}$.
Answer: $\ell(x,y,z)=-x+y-z$.
Answer: $\ell(x,y,z)=-x+y-z$.
Task C. Why does the hollow triangle have $\dim H^1=1$?
Answer: There are three edge cochains, no $2$-simplices, and $\operatorname{rank}D_1=2$, so $\dim H^1=3-2=1$.
Answer: There are three edge cochains, no $2$-simplices, and $\operatorname{rank}D_1=2$, so $\dim H^1=3-2=1$.
Task D. Why does the filled triangle have $\dim H^1=0$?
Answer: The filled face adds $D_2$ with rank $1$, so $\dim H^1=3-2-1=0$.
Answer: The filled face adds $D_2$ with rank $1$, so $\dim H^1=3-2-1=0$.