Let $\pi \colon E \to M$ be a smooth vector bundle. A Riemannian metric on $E$ can be regarded as a global section of the vector bundle $(E\otimes E)^{\ast}$, or more specifically, the subbundle $(S^2E)^{\ast} \subset (E\otimes E)^{\ast}$. However, not every global section $s$ corresponds to a Riemannian metric, as there is no guarantee that $s(x) : S^2E_x \to \mathbb{R}$, when viewed as a symmetric map $E_x \times E_x \to \mathbb{R}$, is positive-definite. So my question is whether we can form a bundle such that positive-definiteness is automatic. More succinctly:

Is there a vector bundle $F$ associated to $E$ such that the global sections of $F$ are precisely the Riemannian metrics on $E$?

If so, can we characterise all Hermitian metrics on a complex vector bundle in a similar way?

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    $\begingroup$ Then the zero section of F would give a canonical choice of metric on E, which does not seem reasonable. $\endgroup$ – Johannes Nordström Jul 13 '12 at 8:49
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    $\begingroup$ The positive-definite metrics are sections of a bundle of convex cones that lives inside $(S^2E)^*$. I see no natural way of making that into a vector bundle. $\endgroup$ – Paul Reynolds Jul 13 '12 at 9:01
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    $\begingroup$ One way of seeing the difficulty is to note that the difference of two metrics (or negative scalar multiple of one metric) is not a metric in general. So there can be no vector bundle structure compatible with natural tensor operations. $\endgroup$ – Minhyong Kim Jul 13 '12 at 9:19
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    $\begingroup$ In my comment above I was saying pretty much the same thing as Minhyong. The space of positive-definite symmetric bilinear forms on $\mathbb{R}^n$ is not a vector space; you can add them (but not subtract) and you can multiply by positive real numbers (but not non-positive ones). So they form a convex cone, that's all I meant. The same is true of the space of Riemannian metrics on a manifold. $\endgroup$ – Paul Reynolds Jul 13 '12 at 16:23
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    $\begingroup$ ...an explicit description of the bundle of cones being $GL(M)/O_n$ where $GL(M)$ is the bundle of linear frames, just like Peter constructs in his answer (which I have just noticed). $\endgroup$ – Paul Reynolds Jul 13 '12 at 16:28

An orthogonal or hermitian structure on $E$ is a section of fibre bundle (which is not a vector bundle). I will deal with the complex case.

The Lie algebra $\mathfrak{gl}(n)$ decomposes into $$ \mathfrak{gl}(n)\simeq \mathfrak{u}(n)\oplus \textrm{Herm}_n, $$ where $\textrm{Herm}_n$ is the vector space of Hermitian $n\times n$ matrices. The exponential map sends it to the positive-definite hermitian matrices: $$ \exp: \textrm{Herm}_n\to \textrm{Herm}_n^+, $$ which form a convex domain ("cone") . More importantly, this cone is actually $$ \textrm{Herm}_n^+ = GL(n,\mathbb{C})/U(n). $$ To see this, consider the action of $GL(n,\mathbb{C})$ on the hermitian matrices by $(T,h)\mapsto \overline{T}^t h T$. Now, do all of this "fibrewise": if $P$ is the frame bundle of $E$, an hermitian metric is a section of the associated bundle with fibre $GL(n,\mathbb{C})/U(n)$.

More coneptually, a choice of reduction of the structure group of a principal $G$-bundle $P$ to a subgroup $K$ is eqivalent to a choice of section of the associated $G/K$ bundle. The hermitian metric is a reduction of the structure group from $GL(n,\mathbb{C})$ to $U(n)$.


Since this question may be a related to your other question

Hermitian Christoffel Symbols

let me say that if you have a complex manifold $X=(M,I)$ with Riemannian metric $g$ on $M$ (compatible with $I$) you can extend $g$ to the complexified tangent bundle $T_{X,\mathbb{C}}$ either as a $\mathbb{C}$-bilinear pairing (as in Kobayashi & Nomizu), or as a sesquilinear pairing $g_{\mathbb{C}}$: see section 1.2 of Huybrechts, "Complex geometry". Now, $T_{X,\mathbb{C}}\simeq T^{1,0}\oplus T^{0,1}$. With the former choice $T^{p,q}$ are isotropic sub-bundles, and only the off-diagonal pairing is non-trivial. With the latter choice (Huybrechts, Griffiths & Harris), the two subbundles are orthogonal and $\left. g_{\mathbb{C}}\right|_{T^{1,0}}$ is $\frac{1}{2}h$, $h= g-i\omega$ (and the conjugate of that on $T^{0,1}$). This turns $E = T^{1,0}$ into an hermitian vector bundle.

In indices, in the first case you have $h_{ab}=0= h_{\overline{a}\overline{b}}$, $h_{a\overline{b}}\neq 0$, while in the second the other way around.

  • $\begingroup$ Thanks Peter. Your aside is also very helpful. I only recently figured out that there were two different ways of approaching the extension to the complexification $\endgroup$ – Michael Albanese Jul 16 '12 at 6:41

I agree with the answer given by Peter Dalakov, and with the comments made so far. On the other hand, there is a solution as close as possible to the requirement.

We consider a default Riemannian metric $g_0$ on the vector bundle $E$. This is the special metric mentioned by Johannes Nordström in his comment. Any other metric can be obtained from this one, by applying a section from the bundle $GL(E)$ (having as fiber at $p\in M$ the general linear group of $E_p$). Any section of $GL(E)$, when applied to $g_0$, will give another Riemannian metric. Also, any section of $GL(E)$ can be written as $\exp(s)$, where $s$ is a section of the vector bundle $SL(E)$.

Hence, we can take as the desired vector bundle the bundle $SL(E)$. Any section $s$ of it will provide a transformation $\exp(s)$ which, when applied to the specially chosen metric $g_0$, gives another metric. Any metric can be obtained this way.

Note 1:

This works similarly for the Hermitian case.

Note 2:

If for the particular problem is necessary to work with generic linear combinations of the metrics themselves, then the solution proposed here will not be of use. Instead, one needs geometric methods which apply to metrics having variable signature, hence also being degenerate. Such methods were developed in arXiv:1105.0201, arXiv:1105.3404, and arXiv:1111.0646. One can define for example covariant derivatives for special tensor fields and differential forms, and also define the Riemann tensor $R_{abcd}$, although these constructions are apparently forbidden by $g_{ab}$ not being always invertible.

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    $\begingroup$ If $SL(E)$ is the associated vector bundle to the adjoint representation of $\mathfrak{sl}_n$, then different sections of $SL(E)$ can give the same metric in this way right? So by 'as close as possible' you mean, it's a bit bigger than we'd ideally want? $\endgroup$ – Paul Reynolds Jul 13 '12 at 16:58
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    $\begingroup$ It is bigger, more sections will give the same metric. They are related by sections of the orthogonal bundle. Factoring over the orthogonal bundle will reduce it, but the result is not a vector bundle. $\endgroup$ – Cristi Stoica Jul 13 '12 at 19:34
  • $\begingroup$ Thanks Cristi. After the first couple of comments on this question, I knew the answer to my question was no, but it interesting that you can get pretty close. $\endgroup$ – Michael Albanese Jul 16 '12 at 6:42

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