Classifying Octonionic-Linear Operators by Alexander A. Putnam AN EXPOSITORY PAPER submitted to Oregon State University In partial fulfillment of the requirements for the degree of Master of Science in Mathematics Presented May 26, 2017 Commencement June 2017 Abstract The goal of this paper is to classify linear operators with octonionic coefficients and octonionic variables. While building up to the octonions we also classify linear operators over the quaternions and show how to relate the linear operators over the quaternions and octonions to matrices. We also construct a basis of linear operators that maps to the canonical basis of matrices for each space. Finally, we discuss automorphisms of the octonions, a special subset of the linear operators. 1 Contents 1 A Review of Division Algebras 3 2 Real, Complex and Quaternionic Linear Operators 7 3 Octonionic Linear Operators 14 4 Relating Matrices to Octonionic Linear Operators 29 5 Automorphisms of the Octonions 38 6 Summary 39 7 References 40 2 1 A Review of Division Algebras Linear operators are a very interesting area in mathematics. They can be used as an introduction for students first learning about functions and transformations, and also for advanced mathemati- cians trying to simplify complicated problems. We use linear operators not only to help discover the structure of the space they live in but also to build up to more complicated functions and even to show that two spaces are isomorphic. So when exploring a new number system, such as the octonions, a great place to start is with the linear operators. Before we begin our discussion of linear operators over the octonions, we will introduce some basic properties of the octonions. To understand the basics of the octonions, it is helpful to start with the real numbers and build larger division algebras up to the octonions. So, we first define a division algebra. A division algebra over a field F, is a vector space over F, with a bilinear operation, ∗, denoted by (V; F; ∗), such that the following properties hold: 1. For all v; w 2 V; v ∗ w 2 V . 2. For all α 2 F and any v; w 2 V ,(αv) ∗ w = α(v ∗ w) = v ∗ (αw). 3. There exists an element e 2 V , such that for all v 2 V; e ∗ v = v ∗ e = v. 4. For all nonzero v 2 V , there exist elements u; w 2 V , such that u ∗ v = v ∗ w = e. 5. For all v; w; x 2 V , v ∗ (w + x) = v ∗ w + v ∗ x and (w + x) ∗ v = w ∗ v + x ∗ v. It is easy to see that if a space, V , is a vector space over a field of scalars, W , with V ⊆ W , then V is a division algebra, because we can define the vector multiplication and division by the multiplication and division in the field. Therefore, the real numbers over R is a division algebra. The next vector space to consider is the complex numbers, C, over R. We can describe the complex numbers as R ⊕ Ri, a direct sum of the real numbers with the real numbers multiplied by an imaginary number i such that i2 = −1. So, elements of C have the form a + bi, where a and b are real numbers. Because our scalars are only real numbers, it is not trivial that C is a division algebra. But, it is not very hard to check that C is a division algebra. If we try this again with a new imaginary unit j we get R⊕Ri⊕Rj, a three dimensional vector space. Now our elements have the form a + bi + cj and, as with i, j2 = −1. We now check if this space is closed under multiplication, (a + bi + cj)(x + yi + zj) = ax + ayi + azj + bix + biyi + bizj + cjx + cjyi + cjzj; (1.1) and in this product we have terms with ij. When Hamilton was first trying to construct the quaternions in 1843, he attempted to create a three-dimensional division algebra with no success. Hamilton realized that no such division algebra exists, and that an extra dimension was needed to close the space. One example of how an attempt to create a three-dimensional algebra fails is as follows: we can try to define ij = −1, so that we get only three terms again. But, this extra relation on i and j allows us to make the following computation, ij = −1 =) i(ij) = i(−1) =) −j = −i =) i = j: (1.2) 3 If i = j, then we have collapsed our space back into the complex plane. The quaternions, H, are a vector space over R and are expressed as R⊕Ri⊕Rj ⊕Rij = C⊕Cj. We say that ij = −ji (i.e. we say i and j anti-commute), then (ij)(ij) = −(ij)(ji) = i2 = −1. For simplicity we will define ij := k. Because i and j anti-commute, in general elements of the quaternions do not commute, which means we can find two elements of the quaternions, a; b, such that ab 6= ba. For example, consider the elements, 1 + i and 2 − j. Then, (1 + i)(2 − j) = 2 − j + 2i + i(−j) = 2 + 2i − j − k; (1.3) but, (2 − j)(1 + i) = 2 + 2i − j(−j)i = 2 + 2i − j + ij = 2 + 2i − j + k: (1.4) So, we need to be careful when we multiply elements of the quaternions not to switch the order of the terms. The quaternions are a four-dimensional division algebra and a proof of this can be found in [1]. Finally, we can describe the octonions. If we try to extend the quaternions to a larger division algebra then, as before, if we add a new unit, `, we also need to add every combination of units, i`; j` and k`. Thus, to create the octonions, we need to consider O = H ⊕ H`. Therefore, each element of our new space has the form a1 +a2i+a3j+a4k+a5i`+a6j`+a7k`+a8` where ai 2 R and so the octonions are an 8-dimensional vector space over R. A proof that the octonions are a division algebra can be found in [1]. Because of the size of Table 1.1, the multiplication table, remembering which products have positive sign and which have negative sign can be very challenging. To help remember, we use the 7-point Fano plane instead of the multiplication table. Each node in Figure 1 describes one of the octonionic units and each line then describes a quaternionic subalgebra of the octonions, (fi; j; kg or fk; `; k`g) and so on. If we are going to multiply two elements, a and b, of the set fi; j; k; i`; j`; k`; `g, then their product is the third node along the line connecting the nodes a and b. If, when we go from a to b, we are moving in the direction of the arrow, then the sign is positive. If we go against the arrow, then the sign is negative. For example, consider the line in the middle that connects the units i; j; k. We know, from the definition of the quaternions, that ij = k: (1.5) ∗ 1 i j k i` j` k` ` 1 1 i j k i` j` k` ` i i −1 k −j −` −k` j` i` j j −k −1 i k` −` −i` j` k k j −i −1 −j` i` −` k` i` i` ` −k` j` −1 −k j −i j` j` k` ` −i` k −1 −i −j k` k` −j` i` ` −j i −1 −k ` ` −i` −j` −k` i j k −1 Table 1.1: Multiplication table of octonionic units. 4 If we start at the node representing i, and move along the line to j, then the third element of that line, k, is the product of the two units. Also, because the arrow on the line is pointing in the direction we traveled, we get positive k. Conversely, if we started at j, and moved to i, then we would go against the arrow, and thus get −k. So, the diagram shows us that ji = −k: (1.6) For each of the other lines in the Fano plane, the process described also works, but you have to imagine that the lines extend behind the picture to make circles. As above, if we start at k and then move along the line to `, we are following the arrow to k`. Therefore, k(`) = k`: (1.7) Similarly, if we start at ` and move to k`, then we wrap around behind the picture back to k. Because we are still moving with the arrow, and not against it, the sign is still positive. `(k`) = k: (1.8) Finally, when we start at k`, and move around the back again, to k, we end at ` following the arrow so k`(k) = `: (1.9) However, if we were to switch the order we multiplied the elements in, (`)k, instead of k(`), (k`)` instead of `(k`) or k(k`) instead of k`(k), then our product would be the same unit, k`, k, ` respectively, but it would be negative instead of positive because, in each case, we would now be moving against the arrow, instead of with it. One very interesting consequence is that if you take Figure 1: Octonionic multiplication in the Fano plane. 5 3 units, not in a quaternionic subalgbra (not all on the same line), then they anti-assoiciate. Thus, if you consider the product (k`)(j(`)) = k`(j`) = i: (1.10) But, if we reassociate the terms then ((k`)j)` = (i`)` = −i: (1.11) Therefore, in general, when we multiply two octonions, both the order and the parentheses matter.