# Stuff which squares to $-1$ in the quaternions, thinking geometrically.

How can we think of the set$$\{x \in \mathbb{H} : x^2 = -1\}$$geometrically? Is this set finite or infinite? Are there some more geometric ways of thinking about than meets the eye? Here, $\mathbb{H}$ denotes the quaternions.

• – Kimball Sep 21 '16 at 12:56

First write $x=a+bi+cj+dk$ and compute that $$x^2=a^2+2abi+2acj+2adk-b^2-c^2-d^2.$$ This yields the equations \begin{align}a^2-b^2-c^2-d^2&=-1\\ab&=0\\ac&=0\\ad&=0.\end{align} If $a\neq 0$ then $b=c=d=0$ and $a^2=-1$, which is impossible. Hence, $a=0$ and $b^2+c^2+d^2=1$. Hence, there are infinitely many elements $x\in\mathbb{H}$ satisfying $x^2=-1$. This set can be characterized as the unit sphere in the hyperplane spanned by $i,j,k$.
• @DavidHill I don't follow with your sentence "This set can be characterized as the unit sphere in the hyperplane spanned by $i$, $j$, $k$."... is it possible you can include a bit more detail for the beginners in the audience like myself? – user203482 Sep 20 '16 at 17:42
• @user2503162 think about the equation $b^2+c^2+d^2=1$. What shape does that equation define? – David Hill Sep 20 '16 at 17:45
• That's the unit sphere. I just don't understand the "in the hyperplane spanned by $i$, $j$, $k$" part. Could you elaborate on that? – user203482 Sep 20 '16 at 17:54
• @user2503162 If I choose coordinates $1=(1,0,0,0)$, $i=(0,1,0,0)$, $j=(0,0,1,0)$ and $k=(0,0,0,1)$, then elements that square to $-1$ are of the form $(0,b,c,d)$ with $b^2+c^2+d^2=1$. That is the unit sphere in the subspace spanned by $i$, $j$, and $k$. – David Hill Sep 20 '16 at 17:56