At the beginning of the paper "The curvature of 4-dimensional Einstein spaces", by Singer and Thorpe, the authors define the space $\mathcal{R}$ of curvature tensors of the vector space $V$ as the set of symmetric bilinear transformations on the space of bivectors $\Lambda^{2}(V).$

Few lines after this, they define the ''Bianchi map'' $b$ as an operator $b : \mathcal{R} \rightarrow \mathcal{R}$ in the following way: $$[b(R)](u_{1},u_{2})u_{3}=\sum_{\sigma \in S_{n}} R(u_{\sigma(1)},u_{\sigma(2)})u_{\sigma(3)},$$ but they do not explain the notation. If $R$ is a transformation $$R: \Lambda^{2}(V) \rightarrow \Lambda^{2}(V),$$ and $u_{1},u_{2}$ and $u_{3}$ are, I presume,vectors of $V,$ what is the meaning of $R(u_{\sigma(1)},u_{\sigma(2)})u_{\sigma(3)}$?

Also, they define immediately after this the ''Ricci contraction'' $r$ as an operator from $\mathcal{R}$ to the space of symmetric linear transformations of $V$ by means of: $$\langle r(R)(v),w \rangle=\mathrm{Tr}\{u \rightarrow R(v,u)w \}.$$

I see that this strongly resembles to the Ricci tensor that one usually meets in riemannian geometry, but I have a similar notational problem with this last definition.

If someone could possible clarify the notation and explain a little this way to look at the curvature tensor (or give me some references) I would be really grateful.


Well, Singer and Thorpe do explain the notation (see p.356 in the original text) by saying that

Using the usual isomorphisms defined by the inner product, a curvature tensor on $V$ may be regarded as a $2$-form on $V$ with values in the vector space of skew symmetric endomorphisms of $V$.

So the meaning of $R(x,y)z$ is an endomorphism $R(x,y)$ applied to an element $z$.

By the "usual isomorphisms" they mean "the musical isomorphisms", that is the identification of the tangent and cotangent space.

In the abstract index notation this can be explained as viewing a tensor with symmetries $$ R_{abcd} = R_{[ab][cd]}=R_{[cd][ab]} $$ as equivalent (up to raising an index) to a tensor $R_{ab}{}^c{}_d = R_{[ab]}{}^c{}_d$, and $(R(x,y)z)^c = x^a y^b R_{a b}{}^c{}_d z^d$.

  • $\begingroup$ Thanks for the explanation $\endgroup$ – Qwertuy May 20 '15 at 10:20

Rather than convert the bivector $R(x \wedge y)$ to a 2-form, one can use clifford algebra instead. Define a product operation on arbitrary vectors $a, b, c$ such that

$$aa = g(a,a), \quad (ab)c = a(bc)$$

I use $g$ here for the Riemannian metric instead of $\langle,\rangle$, for reasons that will become clear shortly.

This "geometric product" produces a ring, whose elements are called "multivectors". The exterior algebra is quite similar: it just imposes that $aa = 0$ instead. For this reason, the clifford algebra is a graded vector space as well. Given a multivector $A$, the quantity $\langle A \rangle_k$ denotes the "grade-$k$" part of this multivector. A multivector may have components of several different grades, but often, we deal primarily with objects of single grades.

With this in mind, interpret $R(x\wedge y)z$, for vectors $x, y, z$ using the clifford product as

$$R(x \wedge y)z \equiv \langle R(\langle x y\rangle_2) z\rangle_3 \quad (\text{clifford products})$$

Or, use the following shorthands: $\langle \langle A \rangle_p \langle B \rangle_q \rangle_{p+q} \equiv A \wedge B$ and $\langle \langle A \rangle_p \langle B \rangle_q \rangle_{|p-q|} = A \cdot B$. These shorthands are common, and they let us get rid of a bunch of angled brackets to instead write

$$R(x \wedge y)z \equiv R(x \wedge y) \cdot z \quad (\text{clifford shorthand})$$

It should be clear, however, that regardless of the notation, the operation is well-defined. Merely interpret the Riemann curvature $R$ as a function of bivectors in the clifford algebra to bivectors in the clifford algebra.

Having a product operation that incorporates the Riemannian metric and the wedge products of exterior algebra allows us to avoid converting the bivector into a 2-form.

For matters of practical computation, one might write $R(x \wedge y)$ in terms of some orthogonal basis and use the corollary that, for two orthogonal vectors $u, v$, $uv = -vu$ under the clifford product. This allows us to write $x \wedge y$ in terms of a linear combination of such terms, enabling the use of associativity, or a host of identities can be derived from these terms (for instance, given two vectors $m,n$ and the vector $z$, see that

$$(m \wedge n) \cdot z = (n \cdot z) m - (m \cdot z) n = g(n,z) m - g(m,z) n$$

which is particularly useful here.)

  • $\begingroup$ Thank you. It's all pretty clear now. $\endgroup$ – Qwertuy May 20 '15 at 10:20

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