I am reading Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities by Emmanuel Hebey and he stated on page $22$:

Let $M$ be a compact manifold endowed with two Riemannian metrics $g$ and $\tilde{g}$. As one can easily check, there exists $C > 1$ such that $$\frac{1}{C} g \leq \tilde{g} \leq C g$$ on $M$, where such inequalities have to be understood in the sense of the bilinear forms.

I would like to help to prove this, because I can not give a satisfactory proof with my attempt, but I put it below to show my effort. I also would like to apologize if my proof is very detailed, but I would like to see if I understood very well the argument and what hypothesis are used and how they are used.

It is sufficient to prove that $\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i$ on $M$ for some constant $C > 1$. Suppose that $\tilde{g}$ is a Riemannian metric which is geodesic normal coordinates at $p$ without loss of generality because if the inequalities above are proved, then the inequalities are true for the metric $\tilde{g}$ which is not geodesic normal coordinates at $p$ only changing $C$ by $\frac{C}{A}$, where $A$ denotes the Jacobian of the change of the coordinates. Now, consider $M$ connected (the author assumes in the beginning of the book that manifolds are connected, I think this is used here to define the next metric on $M$) and endowed with the metric $d(p,q) := \inf \left\{ l(\alpha) \ ; \ \alpha \ \text{is a piecewise differentiable curve joining} \ p \ \text{to} \ q \right\}$. Recall that the Riemannian metric $\tilde{g}$ is smooth in the sense that the map

\begin{align*} \tilde{g}: (M,d) &\longrightarrow (\mathscr{L}^2(T_pM \times T_pM, \mathbb{R}),||\cdot||_{op})\\ p &\longmapsto \tilde{g}(p) \end{align*}

is smooth ($||\cdot||_{op}$ denotes the operator norm over $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$), in particular, the map above is a continuous map defined over a compact metric space, then it is uniformly continuous. This part I am stuck, but I want to define a norm $||\cdot||$ over the image of the Riemannian metric $\tilde{g}$ in order to, for every $\varepsilon > 0$, there exists $\delta(\tilde{g}) > 0$ such that

$$q \in B_{\delta(\tilde{g})}(p) \Longrightarrow |\tilde{g}_{ij}(q) - \tilde{g}_{ij}(p)| \leq = ||\tilde{g}(q) - \tilde{g}(p)|| < \varepsilon$$

Choosing $C > 1$ and $\varepsilon := \frac{1}{2} \left( C - \frac{1}{C} \right)$, we have

$$\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i \ (1)$$

on $B_{\delta(\tilde{g})}(p)$ for each $p \in M$.

I do not sure how to do this, once that $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$ and the coordinate fields vary with $p$, therefore I think I can not take simply the operator norm of this space to be $||\cdot||$, but if I can overcome this difficult, then we can do an analogous reasoning for $g$ to obtain

$$\frac{1}{C} \delta_j^i \leq g_{ij} \leq C \delta_j^i \ (2)$$

on $B_{\delta(g)}(p)$ for each $p \in M$.

Defining $\delta := \min \{ \delta(\tilde{g}), \delta(g) \}$, $(1)$ and $(2)$ hold on $B_{\delta}(p)$ for each $p \in M$. Combining $(1)$ and $(2)$ and observing that $\{ B_{\delta}(p) \ ; \ p \in M \}$ is an cover for $M$, we proved the inequalities desired.


We know that

$$\frac{1}{A} g_p(v,v) \leq \tilde{g}_p(v,v) \leq A g_p(v,v) \ (\star)$$

for all $v \in T_pM$ based on what DIdier_ proved. Analogously,

$$\frac{1}{B} \tilde{g}_p(v,v) \leq g_p(v,v) \leq B \tilde{g}_p(v,v) \ (\star \star)$$

for all $v \in T_pM$.

I will try to prove that

$$\frac{1}{C} g_p(u,v) \leq \tilde{g}_p(u,v) \leq C g_p(u,v)$$

for all $u,v \in T_pM$.

Let $q_{g_p}(v) := g_p(v,v)$ and $q_{\tilde{g}_p}(v) := \tilde{g}_p(v,v)$ be the quadratic forms associated to the $g_p$ and $\tilde{g}_p$ respectively, then

$$g_p(u,v) = \frac{q_{g_p}(u+v) - q_{g_p}(u) - q_{g_p}(v)}{2} \ \text{and} \ \tilde{g}_p(u,v) = \frac{q_{\tilde{g}_p}(u+v) - q_{\tilde{g}_p}(u) - q_{\tilde{g}_p}(v)}{2}.$$

This, $(\star)$ and $(\star \star)$ imply that

$$\tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$


$$g_p(u,v) \leq \left( B - \frac{1}{B} \right) \tilde{g}_p(u,v)$$

for all $u,v \in T_pM$, therefore

$$\frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$

for all $u,v \in T_pM$.

Choosing $C > 1$ sufficiently large such that

$$\frac{1}{C} g_p(u,v) \leq \frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v) \leq C g_p(u,v)$$

for all $u,v \in T_pM$ gives the result.

  • 4
    $\begingroup$ see $g$ and $\tilde g$ as functions on $TM$, take the submanifold $S\subset TM$ given by $g=1$, $S$ is compact, let $C$ be the minimum of $\tilde g$ on $S$, then $\tilde g\geq C g$. Now repeat with the maximum. $\endgroup$
    – user8268
    Jun 15, 2020 at 20:22
  • $\begingroup$ Just wanted to note that the expression $\delta^i_j$ is ill-defined on a Riemannian manifold. It does make sense with respect to local coordinates or a local frame of tangent vectors but not globally. $\endgroup$
    – Deane
    Jan 27, 2021 at 20:39

2 Answers 2


You can prove this in a more direct way. It looks like the proof that in a finite dimensional vector space, all norms are equivalent.

Let $S_gM$ be the unit sphere bundle of $(M,g)$, that is $S_gM = \{ (p,v)\in TM | g_p(v,v)=1 \}$. If $M$ is compact, then $S_gM$ is compact too. The smooth function $f$ on $TM$ defined by $f(p,v)= \tilde{g}_p(v,v)$ is then continuous restricted to $S_gM \subset TM$. Notice $f$ is positive, as every $v\in S_gM$ is non-zero. By compactness, there exist $m,M >0$ such that $m\leqslant f(p,v) \leqslant M$ on $S_gM$. You can chose some constant $C>1$ such that $\frac{1}{C} \leqslant m \leqslant M \leqslant C$, so that on $S_gM$, $\frac{1}{C} \leqslant \tilde{g}_p(v,v)\leqslant C$. By the very definition of $S_gM$, we have that for every $(p,v)\in S_gM$, $$\frac{1}{C}g_p(v,v)\leqslant \tilde{g}_p(v,v) \leqslant Cg_p(v,v).$$ Now, the homogeneity of quadratic forms shows that this inequality is true on all of $TM$.

  • $\begingroup$ thanks for your response! By homogeneity of quadratic forms, do you want to say the correspondence between quadratic forms and bilinear forms given in the item $4$ of the definition $9.1$ of this lecture notes? I can not see why your last statement is true $\endgroup$
    – George
    Jun 15, 2020 at 22:16
  • $\begingroup$ I mean if $q$ is a quadratic form, then $q(\lambda v) = \lambda^2 q(v)$. If $v$ is any tangent vector, write $v = \lambda v_0$ with $g_p(v_0,v_0)=1$ and by positivity of $\lambda^2$, the inequalities remain true for $v$. $\endgroup$
    – Didier
    Jun 16, 2020 at 7:54
  • $\begingroup$ Thanks for the clarification, but how to prove that $\frac{1}{C} g_p(u,v) \leq \tilde{g}_p(u,v) \leq C g_p(u,v)$ from $\frac{1}{C} g_p(v,v) \leq \tilde{g}_p(v,v) \leq C g_p(v,v)$? I was thinking in the inequalities that I want to prove while I am writing this comment and I realized that maybe I do not want to prove that $\frac{1}{C} g_p(u,v) \leq \tilde{g}_p(u,v) \leq C g_p(u,v)$, but that $\frac{1}{C} ||g_p|| \leq ||\tilde{g}_p|| \leq C ||g_p||$ in the sense of the operator norms, your answer would be clear if it is that. Is this the author meant with $\frac{1}{C} g \leq \tilde{g} \leq C g$? $\endgroup$
    – George
    Jun 16, 2020 at 15:00
  • $\begingroup$ The inequality I wrote is between quadratic forms. It is what is implicitly asked to prove. And I think this is what is stated when you write $\frac{1}{C}\delta_{ij} \leqslant \tilde{g}_{ij}$ etc. because it is what is relevant for the metric. In general, when it is asked to prove that for two metrics you have $g \leqslant h$ it has to be understood has "the quadratic form $h-g$ is positive". $\endgroup$
    – Didier
    Jun 16, 2020 at 18:50
  • 1
    $\begingroup$ Yes. In fact, the really interesting part in this is the associated norm, because it gives you a way to measure your tangent vectors, and then the smooth functions on $M$. Having equivalent norms on all the tangent spaces can give you (with compactness assumptions) many equivalence results in some good spaces associated, like the Sobolev spaces $\endgroup$
    – Didier
    Jun 16, 2020 at 19:23

The point of this answer is to explain what the question is; the other answer has a perfect proof. As stated in your first quote, $\frac{1}{C} g \leq \tilde{g} \leq C g$ is to be understood in the sense of quadratic forms. This means that for all $x\in M$ and $v\in T_xM$ we have $$ \frac{1}{C} g_x(v,v) \leq \tilde{g}_x(v,v) \leq C g_x(v,v) $$ or equivalently in local coordinates $$ \frac{1}{C} \sum_{i,j}g_{ij}(x)v^iv^j \leq \sum_{i,j}\tilde{g}_{ij}(x)v^iv^j \leq C \sum_{i,j}g_{ij}(x)v^iv^j. $$ This means that the two norms on every $T_xM$ are bi-Lipschitz equivalent and the constant is independent of $x$.

Even when we write $\frac{1}{C} g_{ij} \leq \tilde{g}_{ij} \leq C g_{ij}$, it can be a shorthand for the inequalities in the sense of quadratic forms. That is indeed a much more likely interpretation than a componentwise result.

To underline the importance of working with quadratic forms rather than individual components, let me define three (partial) orders for symmetric square matrices:

  • In the sense of quadratic forms: $A\leq_{qf}B$ means that $v^TAv\leq v^TBv$ for all $v$.
  • Componentwise: $A\leq_{cw}B$ means that $A_{ij}\leq B_{ij}$ for all indices.
  • For all pairs: $A\leq_{p}B$ means that $u^TAv\leq u^TBv$ for all $u$ and $v$.

Now take $$ A = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} $$ and $$ B = \begin{pmatrix} 1&10\\ 10&1 \end{pmatrix}. $$ Clearly $A\leq_{cw}B$, but for $v=(1,-1)$ we have $$ 2 = v^TAv > v^TBv = -18. $$ Thus $A\leq_{cw}B$ does not imply $A\leq_{qf}B$.

In the case of Riemannian metrics proving $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is insufficient, and in general it does not even hold. For example, if $\tilde g$ is the Euclidean metric (the identity matrix) and $g$ is a Riemannian metric with non-zero (perhaps both positive and negative) off-diagonal entries at some point, the componentwise version is false but the version with quadratic forms is still valid.

In general, $A\leq_{p}B$ implies both $A\leq_{qf}B$ (use the same vector twice) and $A\leq_{cw}B$ (choose two basis vectors). While the order given by pairs of vectors implies the correct one, it often fails because because the componentwise one does even though the desired estimate holds true.

What you need is $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$, not $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ or $\frac1Cg\leq_{p}\tilde g\leq_{p}Cg$. Unfortunately your proof that $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$ implies $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is invalid.


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