D O C U M E N T 2 1 3 A U G U S T 1 9 2 1 1 4 5
rel.
th.[2]
Mr.
Born[3]
encouraged me very much to do so. I think he places more
val[ue] on mathematical clarity than on the phys[ics]. Besides, as a recipient I was
not clear about your [physics]. Full comprehension was difficult for me.—
I would also like to make a few additions about the light invariants or the Weyl
invariants. It can be easily shown that the tensor Weyl indicated is the only
Weyl tensor of 2nd order (always excepting the trivial
).[4]
For, first, every
tensor of 2nd order can be expressed by the Riemann tensor and the ’s, which
follows from the representation of in normal coordinates. Second,
,
. (
km
the second extension of ,
i
the
first). Tkm = Tmk. The 20 quantities Ciklm are expressed by the 10 quantities Tkm.
From this follows 10 eqs. for the Ciklm’s. Namely,
gilCiklm
= Ckm = 2Tkm + gkmT.
C =
Ckmgkm
= 6T, 2Tkm = Ckm .
It is [correct that]:
2Ciklm gilCkm gkmCil + gimCkl + gklCim + (gilgkm gimgkl)C = 0. These eqs.
are the sought 10 eqs. Their number is only apparently 20, as the
diminutions[5]
[of
the] 10 expressions are equal to zero, so the true number is equal to 20 10 = 10.
If one substitutes Ciklm,Ckm,C in these eqs. for Riklm,Rkm,R, then the left-hand side
becomes a Weyl tensor,[6] because the C--’s drop away. This is the Hiklm indicated
by Weyl, and it is the only one because these 10 eqs. are the only ones for which
Ciklm exist, because the 10 quantities Tkm are independent of one another. The van-
ishing of Hiklm is not just necessary but also sufficient (contrary to provisional in-
formation by H. Weyl) for the manifold, which can be mapped conformally onto a
Euclidean one. For Hiklm can be written as
Now, Tkm can be determined so that , even when Lkm is arbitrary.
Then , so gik corresponds to a Eu-
clidean manifold.
Hiklm
iklm
g
gik
ds2
Riklm gik Riklm gilTkm gkmTil gimTkl – gklTim – + + Riklm Ciklm + = =
Tkm
1
2
-- -
km
3
4
------
k m
–
1
8
------gkm + =
gkm
6
--------C
1
3
-- -
Hiklm Riklm gilLkm gkmLil gimLkl – gklLim – + + =
Lkm
Rkm
2
-------- - –
gkmR
12
------------
,
+ = Lkm Lmk =
Tkm
-------- - Lkm =
Miklm Riklm
Ciklm
------------ +
Riklm gik-
--------------------------- 0 = = =