Update 9_crystal_structure.md

A very confusing error fixed in this commit "In panel C we show the conventional unit cell with the lattice vectors" should be "In panel D we show the conventional unit cell with the lattice vectors". Two other minor mistakes fixed.

src/9_crystal_structure.md

@@ -360,7 +360,7 @@ We can assign the lattice points to the stars themselves.
 This is a valid choice of a lattice for the periodic structure because each lattice point has the same environment.
 Because the lattice points form triangles, this lattice is called a **triangular lattice**.
 
-The choice of a lattice also defines two linearly independent primitive lattice vectors $\mathbf{a}_2$ and $\mathbf{a}_2$ (panel B):
+The choice of a lattice also defines two linearly independent primitive lattice vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ (panel B):
 $$
 \mathbf{a}_1 = \hat{\mathbf{x}} = \left[ 1, 0\right], \quad \mathbf{a}_2 = \frac{1}{2}\hat{\mathbf{x}} + \frac{\sqrt{3}}{2} \hat{\mathbf{y}}= \left[1/2, \sqrt{3}/2\right].
 $$
@@ -399,7 +399,7 @@ where $f_i$ is the fractional coordinate of $\mathbf{a}_i$ and $N$ is the dimens
 In 2D this equation reduces to
 
 $$
-(f_1, f_2) = f_1 \mathbf{a}_1 + f_1 \mathbf{a}_1.
+(f_1, f_2) = f_1 \mathbf{a}_1 + f_1 \mathbf{a}_2.
 $$
 
 In our case, the basis is $\star(0,0) = 0 \mathbf{a}_1 +0 \mathbf{a}_2$.
@@ -664,7 +664,7 @@ $$
 Bringing $1/\sqrt{3}$ to the other side and dividing both sides by $\sqrt{3}$ yields $f_1 = 2/3$.
 Hence the basis of the second atom is $\mathrm{C}(2/3, 2/3)$.
 
-In panel C we show the conventional unit cell with the lattice vectors
+In panel D we show the conventional unit cell with the lattice vectors
 
 $$
 \mathbf{a}_1 = [\sqrt{3}, 0] a, \quad \mathbf{a}_2 = [0, 3] a.