I introduce elsewhere the polymorphic linear permutation and tracing operator, which I denote τ. Officially, [A,B,C] is the list A←0, B←1, C←2 and τ([1,2,0],[A,B,C]) is the linear map (: a in A, b in B, c in C; c×a×b ← a×b×c :), with similar for other permutations. In practice, the [A,B,C] is generally manifest from context, so I'll write τ[1,2,0] for this, and likewise for other permutations. When some of the entries in τ's (first) argument are not numbers they are symbols and arise in pairs, indicating tracing (a.k.a. contraction) on the relevant tensor factors (which must be dual to one another). Thus τ has, as special cases, the familiar identity operator on any space in the tensor bundle and trace operator on any space of linear automorphisms.
From τ I'll also presume we've introduced antiSym, for which (e.g.) antiSym([R,R]) = (: (a×b −b×a)/2 ← a×b |R⊗R) and, generally,
wherein n! is the factorial of n, defined by 0! = 1, (1+n)! = n!.(1+n);
{iso (n:|n)} is a synonym for {permutations of n}; and sign(s) is the
signature
of the permutation s (obtained from the group homomorphism
({−1,+1}: |{iso (n:|n)}) which maps each transposition to −1 and any composite
of permutations to the product of their signatures).
I now pause to: look briefly at the grand scope of general operators on the tensor bundle; identify a tiny subclass which is orderly enough to be analysed and; then focus attention even closer in on the special subclass of these which I shall call Leibniz operators.
In general, an operator on the tensor bundle is any mapping which delivers a tensor field whenever it is given a tensor field. For example, the mappings
and, for any fixed tensor field t,
which takes any tensor field, u, returning its tensor product with t
are tensor operators. The full general scope of tensor operators, however, is too wild for analysis or for use in analysis: so I focus on these tensor operators having a certain simple relationship with rank.
For any tensor quantity, at any point on the smooth manifold, there is a unique vector space in the tensor algebra of the gradient and tangent spaces in which this tensor quantity lies. This, in turn, is the manifestation at the given point of one rank of our tensor bundle. Tensor fields (officially sections of the tensor bundle) are always obliged to have constant rank (so a section which delivers a tangent at one point only ever delivers a tangent at other points: it never delivers a gradient or any other tensor); and we can only add tensor fields if they have equal rank. I'll denote a rank by either the name of the linear space of tensors of that rank or a list of factors of that space which, combined with bulk(⊗), would yield that linear space. Thus {linear (G:|T)} = G⊗G and [G,G] are synonyms for the rank of linear maps from the tangent bundle, T, to the gradient bundle, G. Note that a list of form ({X}:|n) has n entries and only one possible value, X, for those entries, so is implicitly ({X}: X ←i |n); thus, when I write ({G}:|n) as a rank, it mean the list [G,...,G] of length n.
I say that a tensor operator respects rank precisely if the
rank of its output depends only on the rank of its input: and, furthermore, that
it has (fixed) rank R precisely if its output's rank is just R
tensored with its input's rank. This is true, for instance, of our tensor it
with t
operator, given above – whereas square respects rank without
having fixed rank. [Strictly, I've restricted this to R being tensored on the
left, which one might wish to describe as left-fixed rank to distinguish it from
the obvious analogue, right-fixed rank. However, the theory may all be
developed using only one of these notions: I have chosen to work with left-fixed
rank and find no need to emphasise the chirality.] In support of this, I define
the mapping ({ranks}: rank |{fixed rank tensor operators}) for which rank(Q) is
the rank of Q (i.e. R above).
I shall describe a tensor operator which respects addition, so Q(u+v) = Q(u) + Q(v), as
Before going on to the the subject of Leibniz operators, via some denotations, pause to consider another important property of tensor operators, which expresses locality. I shall say that a tensor operator, Q, is germinal precisely if, for any tensor field f and any open neighbourhood, U, of the smooth manifold, the restriction of Q(f) to U depends only on (:f:U). (I'm sure this is important, though I've yet to find a place for it in the theory.)
In order to consider the action of tensor operators in connection with the
various products
defined on tensor entities, I now introduce a piece of
syntactic sugar to describe a particular combination of a fixed-rank operator with trace and
permutation. For tensor fields u, w of ranks U, W respectively and a tensor
operator Q of (fixed) rank R, I define denotations of form [Q|u*|w], for each
multiplicative
combinator *
that may be defined between u and w:
in each case, Q acts on w, to produce an entity of rank R⊗W, and u then
acts on the W-aspect of this after the usual manner of *
. Specifically:
i.e. apply Q to w, tensor u with the result and shuffle the order to give
it the same rank as Q(u)×w – this is also the rank of Q(u×w).
All the remaining [Q|u*|w] denotations are, in principle, derived from this
(because × is a universal bilinear operator
).
when U and W are dual to one another (to make · defined): thus Q acts on w and then we contract u with the W-aspect of the result. In this case, [Q|u·|w] is actually equal to Q(w)·u, due to symmetry of · on dual spaces; but this construction may sensibly be extended to any u, w for which u·w is defined (not necessarily symmetrically):
when U is H⊗V and W is dual(V)⊗G for some ranks H, G, V of the tensor bundle, so that u·w is in H⊗G and the resulting [Q|u·|w] is in R⊗H⊗G. The case where U and W are dual to one another merely has H= G= scalar fields.
When U is {linear (V::W)} for some V, so that u(w) is a tensor field of rank V, we can rearrange this last case around the standard isomorphism between {linear (V::W)} and V⊗dual(W), so that
is defined by decomposing Q(w) as a sum of terms in R⊗W and allowing u to act on the W-aspect of each, after the fashion of u(w), to yield a term in R⊗V: these terms are then summed to yield the result. While I shall try to avoid this comparatively ugly construction, it does have its place ...
when u is a scalar field, or
when w is a scalar field.
As you might guess, these last two are of more pedagogic than practical relevance: they derive directly from the first case via the fact that the scalar multiplication of u.w is, in both cases, a synonym for u×w.
In manipulating quantities below, I shall make repeated reference to
a basis, b, of
some rank, R. This means a mapping (R:b:) with: (:b|) the
dimension, at each point, of R; R(m) = span(: b(i,m) ←i :) at each point,
m, of the manifold, and; any ({scalar fields}: h :(:b|)) for which sum(:
h(i).b(i) ←i :) = 0 at any point of the manifold has each h(i)=0 at that
point. When using such a basis I shall usually also introduce its dual,
p
, which is a basis of dual(R) with (:p|) = (:b|) satisfying
p(i)·b(i) = 1 and p(i)·b(j) = 0 whenever i and j differ. The
crucial property of these two dual bases is that sum(: b(i)×p(i) ←i
:) is the identity on R, which superficially supposes that (:b|) is finite and
discrete. However, this restriction may (in principle) be dropped if sum
can be understood as the integration associated with some measure on (:b|). I
shall aim to reserve the pair of names b,p for a basis of Q's rank and its dual
(when discussing an operator Q), using the names e,c when the rank is more
arbitrary.
Note: algebraic logicians use the term leibniz operator
to mean something else
altogether. The present discussion is about smooth manifolds, not algebraic
logic.
I shall say that a fixed rank tensor operator, Q,
obeys Leibniz
precisely if its action on products obeys Leibniz's product
rule (here expressed using some denotations given above):
Note that the first of these subsumes multiplication by scalars: the second will shortly emerge as controlling Q's action on trace/permutation operators, including identities.
I shall describe a fixed rank tensor operator, Q, as a Leibniz
operator precisely if it obeys Leibniz and is globally linear. (Note,
2003: I gather that logicians use the term Leibniz operator
for something
else entirely – I'm not going to try to fix this conflict.)
Now observe that, for h a scalar field and Q a fixed rank operator which
respects Leibniz: for any tensor field, u, Q(h.u) = h.Q(u) + Q(h)×u.
Thus, if h and Q satisfy for every tensor field u, Q(h.u) = h.Q(u)
, then
Q(h) must be zero; Q annihilates h. The converse is trivially also true. Thus
a fixed rank tensor operator which obeys Leibniz and respects addition is:
globally linear (hence a Leibniz operator) precisely if it annihilates scalar
global constants and; locally linear precisely if it annihilates all
scalar fields.
A fixed rank tensor operator which respects addition, obeys Leibniz and agrees with d on scalar fields (so its rank is the gradient bundle) is called a differential operator; since d annihilates all globally constant scalar fields, every differential operator is globally linear, hence a Leibniz operator. The tensor fields a differential operator annihilates are described as constant with respect to the differential operator, but one should not confuse this with the notion of constancy we apply to scalar fields (which doesn't depend on choice of differential operator). In general, the only constant field of any tensor rank other than scalar is zero; though some ranks (all with equal numbers of G and T factors) do have, as we'll see below, other natural non-zero constants.
When we have a metric
on our manifold, expressed as a linear map from
tangents to gradients, our attention will be particularly focussed on those
differential operators which regard this linear map as constant. The study of
curvature in such a context is ultimately an analysis of the antisymmetric part
of the square of such a differential operator: this part turns out to be a
Leibniz operator, though (having the same rank as the metric) it is not
differential. Consequently, it proves useful to analyse Leibniz operators in
general and only refer to differential operators as a special case.
In the presence of a local basis, b, of the rank of a Leibniz operator, Q, we can contract Q with the dual, p, of b to obtain a family of scalar-rank Leibniz operators, p·Q, delivering
in which form it is often most easily manipulated. Note that the sum(: b×u×p :) construction is characteristic of the action of the [||] permutation on the identity on rank(Q) in the form sum(: b×p :). Equally, when considering Q's action on a tensor field v on whose rank we have a local basis e with dual c, with u an arbitrary tensor field, we have
Note that both of these results also hold for any fixed rank tensor operator which respects addition and obeys Leibniz: we didn't need the operator to respect global scalings. The same will be true for action on identities and interaction with τ, though I'm only interested in these results for Leibniz operators.
In particular, let transpose(x) be a chart of some neighbourhood in a smooth
manifold M: then each x(i) is a scalar field on M, construed as a co-ordinate,
and dx = ({gradient fields}: d(x(i)) ←i :) is a local basis of the gradient
bundle (and (:x|) = (:dx|) is the dimension
of M). This generates, in
conjunction with its dual (the axis-direction vectors for the chart), bases for
arbitrary ranks of the tensor bundle. For any tensor field u, this gives us a
basis, e, of u's rank, with dual c: these are wholly determined by x. For each
i in (:e|), c(i)·u is a scalar field: it is the i-co-ordinate of u with
respect to e; indeed u = sum(: (c(i)·u).e(i) ←i :).
The natural
differentiation to use, for any given chart, is that
obtained by differentiating the components of tensor fields, implicitly taking
the bases to be constant. This leads me to define
It is readily verified that this does define a differential operator.
It expresses differentiation with respect to the co-ordinates
defined by
the chart transpose(x).
Consider the effect of a Leibniz operator, Q, on the identity linear map, I, on any rank: let u be a tensor field of the rank on which I acts and observe:
Thus Q(I)·u = 0 for any tensor field, u, so Q(I) must be zero.
Thus any Leibniz operator annihilates identity linear maps. This gives us a
non-zero non-scalar that can legitimately be regarded as constant
under all
differential operators. Indeed, as I'll shortly show, all trace/permutation
operators are likewise constant
: identities are merely the easiest case. One
other reason for mentioning identities first, and separately, is that they
suffice to reveal the relationship between a Leibniz operator's actions on dual
pairs of bases.
For any local basis e (with dual c, as before) of a given rank, R, the identity on R's dual is I = sum(: c(i)×e(i) ←i :dim(R)). Apply to this a Leibniz operator, Q:
expand I: then, as Q respects addition and obeys Leibniz,
whence, for each i in (:e|),
whence Q's action on any rank determines its action on the dual of that rank. The formula just given may be re-written as
in which each c.Q(e).c is a scalar and e×c is the basis e and c imply for {linear (R:|R)}. Learn to recognise the scalars c.Q(e).c: they will crop up extensively, particularly (as p.Q(b).p) when R is rank(Q), where they correspond to the Christoffel symbols in the orthodoxy which educated me.
As alluded to above, the constancy of identities is simply a special case of the constancy of all trace/permute operations. First, I show that a simple transposition operator is constant, via
in which (as in the easier case of the identity, above) what we started with appears as one term in the sum with which we end: so the other must be zero. Since this holds for arbitrary u and v, Q(τ[0,1]) must itself be zero. It is not hard to see that the same method may be applied to any transposition and, thus, to an arbitrary permutation. To see that trace is, likewise, a constant action consider:
from which, as before, we conclude that Q(τ[*,*]) must be zero: again, it is not hard to see that this method of reasoning will work for an arbitrary tracing operation. Consequently, all trace/permute operators are annihilated by any Leibniz operator.
Given that
we can now see that any Leibniz operator (e.g. any differential operator) is entirely determined by its actions on scalar and gradient fields.
Consider the action of a locally linear Leibniz operator (so it annihilates all scalar fields) on a tensor field, u, for whose rank we have a local basis, e. Then (using e's dual, c, to obtain the coefficients of u wrt e, which are then scalar fields):
apply Leibniz, noticing that each u·c(i) is a scalar field, hence Q(u·c(i)) = 0, so that each [Q|e(i).|(u·c(i))] is zero: so
which is just the image of u under the linear map which the sum
yields: thus, first, Q acts as a linear map; secondly, the given sum is, in
fact, independent of our choice of basis, e (since any choice of basis must
deliver the same linear map if this is to agree with Q). Consequently, any
locally linear Leibniz operator acts, on each rank, as a linear map: and this
action is determined by the operator's action on gradient fields. (I trust the
reader now sees why I chose to call such operators locally linear
.)
The difference between two Leibniz operators which agree on scalars must,
necessarily, annihilate scalars; that it is Leibniz is trivial. Consequently,
it must be locally linear and, thus, wholly determined by the linear map as
which it acts on gradients. In particular, since all differential operators
agree with d (and thus with each other) on scalar fields, the difference between
a pair of differential operators is always locally linear and determined by the
linear map as which it acts on gradient fields. Thus, given an arbitrary
differential operator (for use as the origin
of the space of possible
differential operators, e.g. the co-ordinate differentiation operator
associated with any chart), there is a direct correspondence between;
In particular, the co-ordinate differentiation operators of two charts necessarily differ by a locally linear gradient-ranked Leibniz operator: d/dx − d/dy is locally linear for any charts x, y. Likewise, any differential operator can be wholly specified, relative to any given co-ordinate differentiation operator, by the the linear map as which the difference between the two acts on gradients.
Suppose we have two Leibniz operators, C and E, of equal rank, R, and a local basis b of their rank, with dual p. Then, for any tensor field u,
which we can apply to a tensor product, in u's place, to find out whether C&on;E is Leibniz:
Now, this differs from respecting Leibniz in precisely the last pair of terms, p·Ev×p·Cw + p·Cv×p·Ew. However, interchanging C and E in these (and swapping the two terms) has exactly the same effect as interchanging i and j – which is to say, a reversal of sign. Thus we find that antiSym([R,R])&on;(C&on;E + E&on;C) obeys Leibniz: since it is manifestly globally linear (because C and E were, and antiSym([R,R]) is locally linear) it is thus a Leibniz operator; i.e., the antisymmetric part of the anti-commutator of two equal-rank Liebniz operators is itself a Liebniz operator.
An alternative approach to the same result is to consider C and E, of equal rank R, and arbitrary tensor fields u and v; first observe that
then use this to infer the action of C&on;E +E&on;C as
in which the first two terms are just what Leibniz would dictate and the second two terms are [0,1]-symmetric. Notice how much more terse the operator-oriented approach is compared to the co-ordinate-based one.
In particular (either way), when C and E are the same operator, we find that the antisymmetric part of the square of any Leibniz operator is a Leibniz operator.
¿ Is there a corresponding result for several Leibniz operators (of equal rank); is their symmetrised product's antisymmetric part necessarily also Leibniz ?
Consider a list ({Leibniz operators of rank R}: Q |n) and the induced average of the composites of permutations of this list, average(: bulk(&on;, Q&on;s) ←(n|s|n) |{monic map}). We can antisymmetrise any output of this in the n factors of R that it gained from applying the n operators. Call the result Z and consider its action on a tensor product u×v.
I shall refer to a Leibniz operator's square's antisymmetric part as the torsion of the given operator:
As shown above, the torsion is also a Leibniz operator: it is locally linear precisely if its action on scalars is zero, in which case I'll describe our original Leibniz operator as torsion-free (notwithstanding that its torsion may be non-zero on other ranks).
In particular, note that the co-ordinate differentiation operator associated with any chart is torsion-free – in fact, the torsion of any chart's co-ordinate differentiation operator is zero (on all ranks). This is rather extreme but provides the basic justification for focusing our attention on torsion-free differential operators while, in general, allowing the torsion to be non-zero on other ranks than {scalar fields}. In principle, there should be a gauge theory around our choice of torsion-free.
Now, consider the action of a torsion-free Leibniz operator, Q: apply its torsion to its rank (rather than to gradient fields), R. Because Q is torsion-free, its torsion acts on its rank as a linear map (R⊗R⊗R: torsion(Q) :R) and thus as a tensor field (which I'll describe as the (generalised) Riemann tensor of the Leibniz operator Q) of rank R⊗R⊗R⊗dual(R); taking its trace yields the Ricci tensor. Various symmetries of these tensors, and their derivatives, are known as the Bianchi identities. I discuss the Riemann tensor and Ricci tensor separately.
For torsion, we could only antisymmetrise over the two contributions to the rank that came from repeated application of our Leibniz operator: for the operator to act on all ranks, we have to accept that we can't make presumptions about the rank of the subject – in particular, we can't presume that it consists (or can be expressed) entirely of rank(Q) factors, so we can't involve the rank of the input to Q&on;Q in the antisymmetrisation. However, if we restrict attention to those ranks which are purely built from rank(Q), we can apply Q once, twice or however many times and then antisymmetrise the entire output, not just its Q-derived aspects.
If we have a Leibniz operator Q of rank R, we can infer an operator, apply
Q then antisymmetrise
, which acts on all ({R}:|n) ranks, with n natural, and on
all sub-ranks of these, such as the antisymmetric subspace of each such rank.
This isn't a Leibniz operator, because it doesn't act on all ranks: but its
action on the ranks on which it does act does respect Leibniz and addition, so
it's very like a Leibniz operator. It turns out to be principally of interest
when (|Q:{scalar fields}) is diverse enough that, at each point of our manifold,
it spans that point's R. This is just the precondition needed to demand that,
for each point of the manifold, there is some list ({scalar fields}:x:), all of
whose outputs are defined in some neighbourhood of our point, for which
(:Q&on;x:) is a local basis of R throughout that neighbourhood. This condition
is (definitively) met when Q is a differential operator but is (emphatically)
not met when Q is the torsion of some torsion-free Leibniz operator.
First note that, if Q is torsion-free, applying Q and then antisymmetrising annihilates every tensor field of form bulk(×, (:Q&on;f:)) for any list ({scalar fields}:f:). It does so because applying Q to such a field yields a sum of terms each of which is a ×-product of items, one of which is Q(Q(f(i))) for some i; since this is symmetric in the two R-factors it provides, the antisymmetrisation stage annihilates it; since this holds for each term, it holds for the entire sum.
When Q is diverse enough
, as above, we can write any tensor field of rank
({R}:|n) as a sum of terms, each of which is f(n).bulk(×,(:Q&on;f|n)) for
some list (:f:1+n) of scalar fields. Since our apply Q then antisymmetrise
operator annihilates the bulk(×,(:Q&on;f|n)) factor, it must map the given
term to wedge(:Q&on;f|1+n). Thus every output of antiSym&on;Q is a sum of
terms, each of which is of form bulk(×, (:Q&on;f:)) for some list, f, of
scalar fields; and each such term is itself annihilated by antiSym&on;Q, from
which we can infer that antiSym&on;Q annihilates all its own outputs.
In particular, we can infer that antiSym({R}:|3)&on;(:Q&on;Q:R) is zero, whence antiSym([R,R,R], Riemann(Q)) = 0, whenever (|Q:{scalar fields}) is sufficiently diverse – e.g. when Q is a differential operator.
Suppose, for some Leibniz operator Q, that we have a linear map, g, from the dual of Q's rank to Q's rank, which is annihilated by Q. The reason for considering this special case is that, when we come to deal with differential operators, we shall be interested in having one which considers the metric of space-time (of which only the symmetric part is of any relevance) to be constant.
What can we deduce ? First (via Q's annihilation of the permutation involved), the linear map's transpose must also be annihilated by Q: whence so are both its symmetric and antisymmetric parts. Consequently, it suffices to consider cases where the linear map is either symmetric or antisymmetric. Now, our linear map may be expressed in terms of its components with respect to a basis; when Q is applied to it in this form, its annihilation implies a relationship between Q's action on our basis and on the scalar fields obtained as g's components:
which promptly tells us Q's action on the components of g in terms of that on the basis as:
Use of a judiciously chosen combination of permutations of this yields an expression from which we'll be able to determine Q's action on the basis:
Consequently, whenever our basis b is mapped to symmetric results by Q, so that Q&on;b = τ[1,0]&on;Q&on;b, we obtain (note that τ[0,1,2] is an identity, so written here solely to show its relationship with the other τ outputs):
whence, if g is antisymmetric, this quantity is zero: if its symmetric part is invertible, we obtain, writing ig for the inverse of (g +τ[1,0](g))/2,
and, for each i,j,k in (:b|),
Thus, knowing that a Leibniz operator annihilates a symmetric isomorphism from its rank's dual to its rank, and having at our disposal a basis of that rank mapped to symmetric results, tells us the action of the Leibniz operator on this basis – i.e. the equivalent of the Christoffel symbols for the operator. Combined with the operator's action on scalar fields, this gives us its action on anything. In the case where the constant tensor field, g, is symmetric and each Q(b(i)) is also symmetric – as happens for a differential operator when each b(i) is the gradient of a co-ordinate scalar field and the constant tensor is the metric – each Q(b(i)) is simply the result of contracting b(i) on the left of
where g\
(pronounced g under
) denotes left-division by g,
just as /2
denotes right-division by 2. When Q is a differential
operator, so that its action on scalar fields is determined separately, this
last tensor quantity suffices to fully determine Q's action on all tensors.
Note that the measure derived from a constant metric is also constant. For the purposes of discussing the measure, let W = Antisym(dim(M), G). This measure, μ, has rank W with k.μ×μ = det(g) for some constant k (it is usual to use ±1, according as det(g) is, in the one-dimensional {linear (W: |Antisym(dim(m), T))}, on the same side of zero as (|w×w ←w :W) or the opposite side). It is note-worthy that, because g is everywhere invertible, μ is nowhere zero. Since W is one-dimensional, any W-rank tensor's derivative is of rank G&tensor;W and so may be expressed as a finite sum of terms, each of which has the form r×w with r of rank G and w of rank W; but W is one-dimensional and μ is nowhere zero, we we can always write w = f.μ for some scalar field f; this lets us re-arrange r×w as f.r×μ. Since all terms in any G&tensor;W-rank tensor can be so rearranged, and we can then factor out the ×μ and add up the left factors to obtain a simple tensor field of rank G, every tensor of this rank is in fact of form r×μ for some tensor field r of rank G. So take Dμ = r×μ and observe that D(μ×μ) = 2.r×μ×μ; aside from a constant factor of k, this is just 2.r×det(g) = D(det(g)), which is necessarily zero; so we can infer that r is in fact zero everywhere. This, in turn, tells us that Dμ is zero and μ is constant.
Maintained by Eddy.