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Section A.2 Sums and Products

In this section we briefly review the properties of direct sums and tensor products of two algebras, \(V\) and \(W\text{.}\)

Subsection A.2.1 Direct Sums

The direct sum \(V\oplus W\) of two vector spaces is just the formal sum of vectors from the two spaces. Thus, the elements of \(V\oplus W\) have the form \(v\oplus w\text{,}\) with \(v\in V\) and \(w\in W\text{.}\) It is straightforward to verify that \(V\oplus W\) is a vector space, assuming some obvious properties for the operation \(\oplus\text{.}\)

The vector \(v\oplus w\) is also written as the ordered pair \((v,w)\text{,}\) from which it is apparent that, for instance, \(\RR\oplus\RR=\RR^2\text{.}\) It is also clear that the direct sum \(V\oplus W\) contains subspaces isomorphic to \(V\) and \(W\text{,}\) namely \(V\oplus0\) and \(0\oplus W\text{,}\) respectively.

To make \(V\oplus W\) into an algebra, we need a product, assuming of course that \(V\) and \(W\) are now algebras themselves. So we define

\begin{equation} (v_1\oplus w_1)(v_2\oplus w_2) = (v_1v_2)\oplus(w_1w_2)\tag{A.2.1} \end{equation}

thus combining the products on the two original algebras into a product on their sum. Notice that

\begin{equation} (v\oplus0)(0\oplus w) = 0\oplus0 = 0\tag{A.2.2} \end{equation}

for any \(v\in V\text{,}\) \(w\in W\text{,}\) which we can summarize as “\(VW=0\)”.

An illustrative example appears in Section 6.7, in which the Clifford algebra \(\Cl(1,0)\) is described both as having basis \(\{\one,\sigma_z\}\) and as being isomorphic to \(\RR\oplus\RR\text{.}\) A natural assumption would be that the two copies of \(\RR\) in the direct sum are spanned by \(\one\) and \(\sigma_z\text{,}\) respectively. This statement is clearly true for the underlying vector spaces, but fails for the corresponding algebras. Why? Because \(\one\sigma_z=\sigma_z\ne0\text{!}\)

The correct interpretation of \(\Cl(1,0)\cong\RR\oplus\RR\) is that there exist two copies of \(\RR\) in \(\Cl(1,0)\) with the necessary properties. A bit of thought leads to the realization that

\begin{equation} (\one+\sigma_z)(\one-\sigma_z) = 0\tag{A.2.3} \end{equation}

so that these two elements can be taken as the basis elements for the two copies of \(\RR\) in the direct sum. Put differently, these two copies of \(\RR\) correspond to the \((1,1)\) and \((2,2)\) components of these \(2\times2\) matrices, rather than to the elements of ranks \(0\) and \(1\) separately.

As also noted in Section 6.7, \(\Cl(1,0)\cong\CC'\text{,}\) under the identification of the matrices \(\one\) and \(\sigma_z\) with the split complex numbers \(1\) and \(L\text{,}\) respectively, with \(L^2=1\text{.}\) Under this identification, the two copies of \(\RR\) in the direct sum are not spanned by \(1\) and \(L\text{,}\) but rather by \(1\pm L\text{,}\) since

\begin{equation} (1+L)(1-L) = 0 .\tag{A.2.4} \end{equation}

Be careful when interpreting directing sums of algebras, which are constrained in ways that direct sums of vector spaces are not.

Subsection A.2.2 Tensor Products

The tensor product of two vector spaces is just the formal product of vectors from the two spaces. Thus, the elements of \(V\otimes W\) are linear combinations of elements of the form \(v\otimes w\text{,}\) with \(v\in V\) and \(w\in W\text{.}\) It is straightforward to verify that \(V\otimes W\) is a vector space, assuming some obvious properties for the operation \(\otimes\text{.}\)

To make \(V\otimes W\) into an algebra, we need a product. So we define

\begin{equation} (v_1\otimes w_1)(v_2\otimes w_2) = (v_1v_2)\otimes(w_1w_2)\tag{A.2.5} \end{equation}

thus combining the products on the two original algebras into a product on their sum. It is again clear that the direct product \(V\otimes W\) contains subspaces isomorphic to \(V\) and \(W\text{,}\) which are now \(V\otimes\one\) and \(\one\otimes W\text{,}\) respectively.

An illustrative example is the tensor product of Pauli matrices, which were introduced in Section 1.3. The simplest case is to consider, say,

\begin{equation} \sigma_z\otimes\sigma_x = \begin{pmatrix} 0\amp1\amp0\amp0\\ 1\amp0\amp0\amp0\\ 0\amp0\amp0\amp-1\\ 0\amp0\amp-1\amp0\\ \end{pmatrix}\tag{A.2.6} \end{equation}

in which copies of \(\sigma_x\) are placed as blocks into a \(4\times4\) matrix, multiplied in each case by the corresponding element of \(\sigma_z\text{.}\) The tensor product is not commutative, since

\begin{equation} \sigma_x\otimes\sigma_z = \begin{pmatrix} 0\amp0\amp1\amp0\\ 0\amp0\amp0\amp-1\\ 1\amp0\amp0\amp0\\ 0\amp-1\amp0\amp0\\ \end{pmatrix}\tag{A.2.7} \end{equation}

where now the blocks are copies of \(\sigma_z\text{,}\) multiplied by the corresponding element of \(\sigma_x\text{.}\) It is worth checking that multiplying these matrices together yields

\begin{equation} (\sigma_z\otimes\sigma_x)(\sigma_x\otimes\sigma_z) = (\sigma_z\sigma_x)\otimes(\sigma_x\sigma_z) = (i\sigma_y)\otimes(-i\sigma_y) = \sigma_y\otimes\sigma_y\tag{A.2.8} \end{equation}

as expected.