What is exponential of a blade? This answer to another question discusses using geometric algebra to find a rotation of an arbitrary angle between two vectors. It involves constructing a versor $R$ from a normalized blade $\hat{B}$ by taking the exponential of it scaled by the angle of rotation: $R=\exp(−\hat{B}θ/2)$.  What is $R$ when expanded, if $\hat{B}=\alpha+ \beta e_1+\gamma e_2+\delta e_1e_2$? 
My guess is it would be similar to $\exp(\alpha x+\beta y) = e^{ax+by}=e^{ax}e^{ay}$, giving $R=e^{\alpha}e^{\beta e_1}e^{\gamma e_2}e^{\delta e_1e_2}$. This doesn't seem correct, though, for two reasons:


*

*the geometric product is not commutative (but addition is) so it seems strange to assume that $a^{b+c}=a^ba^c$

*the result is no longer the sum of products with basis elements $e_1,e_2,e_1e_2$ 



What is the correct way to carry out the exponentiation of a blade?
Edit: I found this answer to a related question just now, and it seems to suggests that $\exp(\hat{B}\theta)=\cos{\theta}+\hat{B}\sin{\theta}$ when $\hat{B}$ is a normalized bivector (or blade).  This suggests to me that, if I correctly set my example blade to $\hat{B}=\alpha e_1e_2+\beta e_1e_3+\gamma e_2e_3$ (in 3D now instead of 2D, so I still have multiple terms), then $R$ becomes the multivector $R=\cos{(-\theta/2)}+\alpha \sin{(-\theta/2)}e_1e_2+\beta \sin{(-\theta/2)}e_1e_3+\gamma \sin{(-\theta/2)}e_2e_3$Is this correct?
 A: Your example:
$$\hat{B} = \alpha + \beta e_1 + \gamma e_2 + \delta e_1 e_2$$
is not a blade, but is a multivector.  Blades are completely antisymmetric sums (i.e. wedge products).  For example, the wedge of two vectors $a, b$ is
$$ a \wedge b = \frac{1}{2} (a b - b a) = \sum_{i < j} \begin{vmatrix} a_i & a_j \\ b_i & b_j \end{vmatrix} e_i e_j $$
In $R^3$ this bivector, a blade, squares to a negative number (the negative of the squared area of the parallogram formed by $a,b$).  Because of this, it's possible to define an exponential of a bivector using power series and it behaves exactly like a complex exponential.
It's not easy to see from the coordinate expansion above, but every wedge product can be expressed as a product of two normal vectors, say $a \wedge b = c d$, where $c \cdot d = 0$.  Since normal vectors anticommute, this means that $B^2 = -c^2 d^2$.  In Euclidean spaces, this is always a negative number, but even generally, this square is always a scalar, so an exponential series representation is simple to construct.
Not every bivector is a blade.  For example in $R^4$, the following grade 2 multivector (i.e. bivector) 
$$B = \alpha(e_1 \wedge e_2) + \beta(e_3 \wedge e_4)$$
cannot be written as a product of normal vectors.  A consequence of this is:
$$e^B \ne e^{\alpha(e_1 \wedge e_2)} e^{\beta(e_3 \wedge e_4)}.$$
Any quantum mechanics book covering spin operators in detail has the formula for $e^{A + B}$ (which I forget off the top of my head).  It has three factors $e^A$, $e^B$, and a cross term.  Only if $A$ and $B$ commute is the exponential of a sum the product of the exponentials.
In your edit, you asked about a normalized bivector in $R^3$ of the form
$$\hat{B} = \alpha e_1 e_2 + \beta e_2 e_3 + \gamma e_3 e_1.\qquad (1)$$
You computed the exponential of that correctly.  Paraphrasing what you wrote, that exponential is:
$$e^{\hat{B} \theta/2} = \cos(\theta/2) + \hat{B} \sin(\theta/2).$$
This is what you wrote, with $\hat{B}$ expanded in coordinates.  In $R^3$ any bivector is a 2-blade, so can be written as a wedge product, or as a product of normal vectors.  An example:
$$\frac{1}{\sqrt{3}}(e_{1} e_{2} +
e_{2} e_{3} +
e_{3} e_{1})
= \frac{ e_1 + e_2 - 2 e_3 }{\sqrt{6}} \frac{ e_2 - e_1 }{\sqrt{2}}
= \frac{ e_3 - e_2 }{\sqrt{2}} \frac{ 2 e_1 - e_2 - e_3 }{\sqrt{6}}$$
These are factorizations of the form $\hat{B} = c d$, where $c \cdot d = 0$, and
$$\hat{B}^2 = c d c d = -d c c d = - c^2 d^2 = -1.$$ 
We see that this normalized $\hat{B}$ behaves just like an imaginary, so the Euler's $cos + i\, sin$ relationship follows.
More generally, to show that any normalized $R^3$ bivector behaves like an imaginary, we can write the bivector as the dual of a normalized vector:
$$\hat{B} = I \hat{n},$$
where $I = e_1 e_2 e_3$ is a unit trivector (also called the $R^3$ pseudoscalar).  For the general $R^3$ $\hat{B}$ in (1) above, this is:
$$\hat{B} = \alpha e_1 e_2 + \beta e_2 e_3 + \gamma e_3 e_1
= 
I (\alpha e_3 + \beta e_1 + \gamma e_2),$$
showing that $\hat{B}$ is dual to the unit vector with direction cosines $(\beta, \gamma, \alpha)$.  You can show that $I$ commutes with all grades, and that $I^2 = -1$, so
$$\hat{B}^2 = I \hat{n} I \hat{n} = I^2 \hat{n}^2 = (-1)(1).$$
Again Euler's $e^{i\theta} = cos(\theta) + i\, sin(\theta)$ relationship follows for $i = \hat{B}$.
