Tensor Calculus – Why the Metric and Levi-Civita Tensor are the Only Invariant Tensors

diffeomorphism-invarianceinvariantslorentz-symmetrytensor-calculus

The only numerical tensors that are invariant under some relevant symmetry group (the Euclidean group in Newtonian mechanics, the Poincare group in special relativity, and the diffeomorphism group in general relativity) are the metric $g_{\mu \nu}$, the inverse metric $g^{\mu \nu}$, the Kronecker delta $\delta^\mu_\nu$, and the Levi-Civita tensor $\sqrt{|\det g_{\mu \nu}|} \epsilon_{\mu \nu \dots}$. (Note that $\delta^\mu_\nu = g^{\mu \rho} g_{\rho \nu}$, so only two of the first three invariant tensors are independent). This result is extremely useful for constructing all possible scalar invariants of a given set of tensor fields, but I've never actually seen it proven. How does one prove that there are no other invariant tensors?

Best Answer

For a proof that the only ${\rm SO}(N)$ invariant tensors are products of $\delta_{ab}$'s and Levi-Civita symbols see M. Spivak, A Comprehensive Introduction to Differential Geometry (second edition) Vol. V, pp. 466-481. The number of pages required for the argument shows that it is not trivial.

I've just looked up the third edition of Spivak vol V. What is needed is theorem 35 on page 327. This is the section entitled "A Smattering of Classical Invariant Theory." He writes in terms of scalar invariants, but of course an invariant tensor becomes a scalar invariant when contracted with enough vectors, and any such scalar invariant arises from an invariant tensor.

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