A Geometric Interpretation of Curvature Inequalities

A Geometric Interpretation of Curvature Inequalities on Hypersurfaces via Ravi Substitutions in the Euclidean Plane Bogdan D. Suceav˘a California State University, Fullerton

Geometry and Topology Seminar, CSUF

Bogdan D. Suceav˘ a California State University, Fullerton

A Geometric Interpretation of Curvature Inequalities on Hypersur

Question: how can we visualize curvature?



Question: how can we visualize curvature?

Consider the arbitrary three real numbers x, y , z > 0. By using the substitutions a = y + z, b = z + x, and c = x + y , → there exists a non-degenerate triangle in the Euclidean plane with sides of lengths a, b, c. Argument: a + b = x + y + 2z > x + y = c, hence the triangle inequality is always satisfied. Bogdan D. Suceav˘ a California State University, Fullerton


The converse of this claim is also true. Namely, if a, b, c are the sides of a triangle in the Euclidean plane, then the system given through the equations a = y + z, b = z + x, and c = x + y , has a unique solution x = s − a, y = s − b, z = s − c, where we denote by s the semiperimeter of the triangle, namely s = 12 (a + b + c). Bogdan D. Suceav˘ a California State University, Fullerton


We will use this elementary remark to establish a correspondence between certain facts holding in triangle geometry and a class of inequalities with curvature invariants for three-dimensional hypersurfaces in the four-dimensional real Euclidean space.



We will use this elementary remark to establish a correspondence between certain facts holding in triangle geometry and a class of inequalities with curvature invariants for three-dimensional hypersurfaces in the four-dimensional real Euclidean space. Some classical facts in advanced Euclidean geometry can be proved by substituting instead of a, b, c the variables x, y , z. This technique is called in some recent references Ravi’s substitutions.



To better see how we use Ravi’s transformation, we list here a few useful formulae for a triangle ∆ABC lying in the Euclidean plane.



To better see how we use Ravi’s transformation, we list here a few useful formulae for a triangle ∆ABC lying in the Euclidean plane. Denote by A the area of the triangle, by R its circumradius, r its inradius, by s its semiperimeter and by P its perimeter. Then Heron’s p formula is p A = s(s − a)(s − b)(s − c) = xyz(x + y + z).



To better see how we use Ravi’s transformation, we list here a few useful formulae for a triangle ∆ABC lying in the Euclidean plane. Denote by A the area of the triangle, by R its circumradius, r its inradius, by s its semiperimeter and by P its perimeter. Then Heron’s p formula is p A = s(s − a)(s − b)(s − c) = xyz(x + y + z). The inradius can be obtained as r A xyz r= = , s x +y +z



To better see how we use Ravi’s transformation, we list here a few useful formulae for a triangle ∆ABC lying in the Euclidean plane. Denote by A the area of the triangle, by R its circumradius, r its inradius, by s its semiperimeter and by P its perimeter. Then Heron’s p formula is p A = s(s − a)(s − b)(s − c) = xyz(x + y + z). The inradius can be obtained as r A xyz r= = , s x +y +z and the circumradius can be obtained as R=

abc (y + z)(z + x)(x + y ) p = . 4A 4 xyz(x + y + z)



Application 1. The isoperimetric inequality for √ a triangle in the Euclidean plane. This inequality states 36A ≤ 3 · P 2 .



Application 1. The isoperimetric inequality for √ a triangle in the Euclidean plane. This inequality states 36A ≤ 3 · P 2 . For a proof, note that √ P = 2s, hence the inequality can also be stated as 9A ≤ 3 · s 2 . By using Ravi substitutions this yields p √ 9 xyz(x + y + z) ≤ 3(x + y + z)2 ,



Application 1. The isoperimetric inequality for √ a triangle in the Euclidean plane. This inequality states 36A ≤ 3 · P 2 . For a proof, note that √ P = 2s, hence the inequality can also be stated as 9A ≤ 3 · s 2 . By using Ravi substitutions this yields p √ 9 xyz(x + y + z) ≤ 3(x + y + z)2 , which by direct computation turns out to be √ 3

xyz ≤

x +y +z , 3

i.e. the AM-GM inequality for three positive real numbers. Equality holds if and only if x = y = z, i.e. when the triangle is equilateral.



To obtain an interpretation in terms of curvature →



To obtain an interpretation in terms of curvature → consider a correspondence between



To obtain an interpretation in terms of curvature → consider a correspondence between a triangle ∆ABC with lengths of sides a, b, c, and corresponding x = s − a, y = s − b, z = s − c, and



To obtain an interpretation in terms of curvature → consider a correspondence between a triangle ∆ABC with lengths of sides a, b, c, and corresponding x = s − a, y = s − b, z = s − c, and a convex three-dimensional smooth hypersurface such that at a point p on the surface the principal curvatures are λ1 (p) = x, λ2 (p) = y , λ3 (p) = z. (The convexity condition is needed to insure that x, y , z > 0 in an open neighborhood around p.)



Application 1.

Ravi transformation of the isoperimetric inequality.



Application 1.

Ravi transformation of the isoperimetric inequality. Note that now x + y + z = 3H, xyz = K .



Application 1.

Ravi transformation of the isoperimetric inequality. Note that now x + y + z = 3H, xyz = K . Then the isopetrimetric inequality becomes √ √ 9 K · 3H ≤ 3(3H)2 , i.e. K ≤ H 3, an inequality between the Gauss-Kronecker curvature and the mean curvature, which is directly equivalent to AM-GM inequality.



Application 2. Mitrinović Inequality in a triangle in the Euclidean plane.



Application 2. Mitrinović Inequality in a triangle in the Euclidean plane. Let ∆ABC be a triangle with inradius r and semiperimeter s. Then √ s ≥ 3 3. r



Application 2. Mitrinović Inequality in a triangle in the Euclidean plane. Let ∆ABC be a triangle with inradius r and semiperimeter s. Then √ s ≥ 3 3. r Proof via Ravi’s technique: √ s s ·s s2 (x + y + z)2 = = =p ≥ 3 3. r s ·r A xyz(x + y + z)



Application 2. Mitrinović Inequality in a triangle in the Euclidean plane. Let ∆ABC be a triangle with inradius r and semiperimeter s. Then √ s ≥ 3 3. r Proof via Ravi’s technique: √ s s ·s s2 (x + y + z)2 = = =p ≥ 3 3. r s ·r A xyz(x + y + z) A direct computation yields the AM-GM inequality for three positive real numbers, that is √ 3

xyz ≤


x +y +z . 3


Application 2. Mitrinović Inequality in a triangle in the Euclidean plane. Let ∆ABC be a triangle with inradius r and semiperimeter s. Then √ s ≥ 3 3. r Proof via Ravi’s technique: √ s s ·s s2 (x + y + z)2 = = =p ≥ 3 3. r s ·r A xyz(x + y + z) A direct computation yields the AM-GM inequality for three positive real numbers, that is √ 3

xyz ≤

x +y +z . 3

Back to 3-dimensional hypersurfaces: √ √ Transforming Mitrinović inequality, we obtain (3H)2 ≥ 3 3 K · 3H, which reduces immediately to K ≤ H 3 . Bogdan D. Suceav˘ a California State University, Fullerton


Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r .



Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r .



Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r . Proof, by using the Ravi substitutions



Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r . Proof, by using the Ravi substitutions we need to show that the following holds: r xyz (y + z)(z + x)(x + y ) p ≥2 , x +y +z 4 xyz(x + y + z)



Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r . Proof, by using the Ravi substitutions we need to show that the following holds: r xyz (y + z)(z + x)(x + y ) p ≥2 , x +y +z 4 xyz(x + y + z) or, by a direct computation: (y + z)(z + x)(x + y ) ≥ 8xyz.



Application 3. Euler’s inequality (obtained originally in 1763). In any triangle in the Euclidean plane, the circumradius and the inradius satisfy R ≥ 2r . Proof, by using the Ravi substitutions we need to show that the following holds: r xyz (y + z)(z + x)(x + y ) p ≥2 , x +y +z 4 xyz(x + y + z) or, by a direct computation: (y + z)(z + x)(x + y ) ≥ 8xyz. √ This last inequality holds true since x + y ≥ 2 xy , and two other similar inequalities, multiplied term by term, conclude the argument. Bogdan D. Suceav˘ a California State University, Fullerton


Application 3. What is the equivalent of Euler’s inequality on 3-dimensional hypersurfaces?



Application 3. What is the equivalent of Euler’s inequality on 3-dimensional hypersurfaces? We remind here a class of curvature invariants introduced recently, the sectional absolute mean curvatures defined by |λi (p)| + |λj (p)| . H¯ij (p) = 2



Application 3. What is the equivalent of Euler’s inequality on 3-dimensional hypersurfaces? We remind here a class of curvature invariants introduced recently, the sectional absolute mean curvatures defined by |λi (p)| + |λj (p)| . H¯ij (p) = 2 By using the same method as above, Euler’s inequality R ≥ 2r inequality transforms into (y + z)(z + x)(x + y ) ≥ 8xyz, which admits the interpretation in terms of curvature invariants as follows: H¯12 · H¯23 · H¯31 ≥ K .



Application 3. What is the equivalent of Euler’s inequality on 3-dimensional hypersurfaces? We remind here a class of curvature invariants introduced recently, the sectional absolute mean curvatures defined by |λi (p)| + |λj (p)| . H¯ij (p) = 2 By using the same method as above, Euler’s inequality R ≥ 2r inequality transforms into (y + z)(z + x)(x + y ) ≥ 8xyz, which admits the interpretation in terms of curvature invariants as follows: H¯12 · H¯23 · H¯31 ≥ K . (This is different than the previous interpretation K ≤ H 3 .) Bogdan D. Suceav˘ a California State University, Fullerton


These ideas are in: A Geometric Interpretation of Curvature Inequalities on Hypersurfaces via Ravi Substitutions in the Euclidean Plane, to appear in the Mathematical Intelligencer.



Theorem Let M 3 ⊂ R4 be a smooth hypersurface and λ1 , λ2 , λ3 be its principal curvatures in the ambient space R4 . Let p ∈ M be an arbitrary point. Denote by H(p) = 13 (λ1 + λ2 + λ3 ) the mean curvature, by scal(p) = sec(e1 ∧ e2 ) + sec(e3 ∧ e1 ) + sec(e2 ∧ e3 ) = λ1 (p)λ2 (p) + λ3 (p)λ1 (p) + λ2 (p)λ3 (p) the scalar curvature, and by K (p) = λ1 (p)λ2 (p)λ3 (p) the Gauss-Kronecker curvature. Then q 2 1 2 H+ −H − Hq + scal ≤ 3 3 3 q 1 2 ≤K ≤ H− −H 2 + Hq + scal , 3 3 3 p √ where q = 9H 2 − 9ρ = 9H 2 − 3 scal. The equality holds if and only if the point is semi-umbilical, i.e. if the following holds (λ1 (p) − λ2 (p))(λ3 (p) − λ1 (p))(λ2 (p) − λ3 (p)) = 0. Bogdan D. Suceav˘ a California State University, Fullerton


Note first that 9H 2 − 9ρ ≥ 0 is a consequence of Chen’s fundamental inequality with classical curvature invariants, so it makes sense to take q 2 = 9H 2 − 9ρ.



Note first that 9H 2 − 9ρ ≥ 0 is a consequence of Chen’s fundamental inequality with classical curvature invariants, so it makes sense to take q 2 = 9H 2 − 9ρ. If q = 0 at p ∈ M, then, by the equality case in Chen’s fundamental inequality with classical curvature invariants, p is an umbilical point. In this case, the double inequality in the statement turns out to be an identity: 2 K = −H 3 + scal · H, 3 which reduces immediately to 2λ3 = 2λ3 .



Note first that 9H 2 − 9ρ ≥ 0 is a consequence of Chen’s fundamental inequality with classical curvature invariants, so it makes sense to take q 2 = 9H 2 − 9ρ. If q = 0 at p ∈ M, then, by the equality case in Chen’s fundamental inequality with classical curvature invariants, p is an umbilical point. In this case, the double inequality in the statement turns out to be an identity: 2 K = −H 3 + scal · H, 3 which reduces immediately to 2λ3 = 2λ3 . Now assume that p is not an umbilical point, i.e. q > 0 at p ∈ M.



Consider the function given by the real polynomial of degree three whose roots at every point at the principal curvatures of the hypersurface: f (x) = (x − λ1 )(x − λ2 )(x − λ3 ) = x 3 − 3Hx 2 + scal · x − K .



Consider the function given by the real polynomial of degree three whose roots at every point at the principal curvatures of the hypersurface: f (x) = (x − λ1 )(x − λ2 )(x − λ3 ) = x 3 − 3Hx 2 + scal · x − K . Then the derivative of f is the quadratic function f 0 (x) = 3x 2 − 6Hx + scal.



Consider the function given by the real polynomial of degree three whose roots at every point at the principal curvatures of the hypersurface: f (x) = (x − λ1 )(x − λ2 )(x − λ3 ) = x 3 − 3Hx 2 + scal · x − K . Then the derivative of f is the quadratic function f 0 (x) = 3x 2 − 6Hx + scal. The equation f 0 (x) = 0 has the real roots: √ q 6H ± 36H 2 − 12 · scal =H± . x1,2 = 6 3



Consider the function given by the real polynomial of degree three whose roots at every point at the principal curvatures of the hypersurface: f (x) = (x − λ1 )(x − λ2 )(x − λ3 ) = x 3 − 3Hx 2 + scal · x − K . Then the derivative of f is the quadratic function f 0 (x) = 3x 2 − 6Hx + scal. The equation f 0 (x) = 0 has the real roots: √ q 6H ± 36H 2 − 12 · scal =H± . x1,2 = 6 3 Denote by x1 = H + q3 , and by x2 = H − q3 , that is x2 ≤ x1 . Remark that the second derivative is f 00 (x) = 6x − 6H. Bogdan D. Suceav˘ a California State University, Fullerton


Remark that the second derivative is f 00 (x) = 6x − 6H, and this yields q − 6H = 2q > 0, f 00 (x1 ) = 6 H + 3 which means that f has a local minimum at x1 . Similarly, q f 00 (x2 ) = 6 H − − 6H = −2q < 0, 3 and this means that the original function f has a local maximum at x2 .



So, f is a polynomial of degree three, with three real roots. This means the local minimum must have a nonpositive value, while the local maximum must have a nonnegative value: 3H + q 3H − q ≥ 0, f ≤ 0. f 3 3 By evaluating f at these two values corresponding to x2 and x1 , respectively, we have:

3H − q 3

3

3H + q 3

3

− 3H − 3H

3H − q 3

2

3H + q 3

2

+ scal ·


+ scal ·

3H − q 3

3H + q 3

− K ≥ 0,

− K ≤ 0.


By joining these two inequalities, we have:

3H + q 3 ≤

3

− 3H

3H + q 3

3

3H − q 3

− 3H

2

+ scal ·

3H − q 3


3H + q 3

2

+ scal ·

≤K ≤

3H − q 3

.


A direct computation shows that the upper bound is: q q 2 q − 3H H − H− H− + scal = 3 3 3 q2 q 2 2 2 H − Hq + − 3H + Hq + scal = = H− 3 3 9 q 9H 2 − 3 · scal 1 2 = H− + scal . −2H + Hq + 3 3 9 This last expression reduces to the right hand side term in the double inequality stated in the Theorem. The expression in the lower bound is obtained by a similar computation.



To complete the proof, we need to discuss now when the equality holds in the two stated inequalities. Equality holds in either of the two inequalities when we have either 3H − q f = 0, 3



To complete the proof, we need to discuss now when the equality holds in the two stated inequalities. Equality holds in either of the two inequalities when we have either 3H − q f = 0, 3 or

f

3H + q 3


= 0.


To complete the proof, we need to discuss now when the equality holds in the two stated inequalities. Equality holds in either of the two inequalities when we have either 3H − q f = 0, 3 or

f

Suppose f

3H−q 3

3H + q 3

= 0.

= 0. Then f (x) = (x − λ1 )(x − λ2 )(x − λ3 ),

3H−q 3

which means that should be one of the roots. Suppose 3H−q = λ1 . Then we get λ2 + λ3 − q = 2λ1 . 3



By squaring both sides in λ2 + λ3 − 2λ1 =

p

9H 2 − 3 · scal

we obtain 3λ21 + 3λ2 λ3 − 3λ1 λ2 − 3λ1 λ3 = 0, which factors as (λ1 − λ2 )(λ1 − λ3 ) = 0, which means that at least a second principal curvature λ2 or λ3 must equal λ1 , and this can be written as (λ1 (p) − λ2 (p))(λ3 (p) − λ1 (p))(λ2 (p) − λ3 (p)) = 0. = 0 yields a similar conclusion. The case f 3H+q 3



This idea is in the paper An estimate for the Gauss-Kronecker curvature of 3-dimensional smooth hypersurfaces in E4 . Int. Electron. J. Geom. 7, No. 2, 1–6 (2014).



Theorem Let M 3 ⊂ R4 be a smooth hypersurface. Let p ∈ M be an arbitrary point and denote by k1 , k2 , k3 be its principal curvatures.



Theorem Let M 3 ⊂ R4 be a smooth hypersurface. Let p ∈ M be an arbitrary point and denote by k1 , k2 , k3 be its principal curvatures. Suppose that |k1 |, |k2 |, |k3 | ∈ [1, 1 + ε].



Theorem Let M 3 ⊂ R4 be a smooth hypersurface. Let p ∈ M be an arbitrary point and denote by k1 , k2 , k3 be its principal curvatures. Suppose that |k1 |, |k2 |, |k3 | ∈ [1, 1 + ε]. Denote by A the amalgamatic curvature, by H the mean curvature, H¯ the absolute mean curvature. Then the following inequalities hold true at every point p ∈ M :



Theorem Let M 3 ⊂ R4 be a smooth hypersurface. Let p ∈ M be an arbitrary point and denote by k1 , k2 , k3 be its principal curvatures. Suppose that |k1 |, |k2 |, |k3 | ∈ [1, 1 + ε]. Denote by A the amalgamatic curvature, by H the mean curvature, H¯ the absolute mean curvature. Then the following inequalities hold true at every point p ∈ M : 2 2 A ≤ H¯ ≤ 5 + [1 + (1 + ε) ] A. 1+ε Equality holds for absolutely umbilical points.



Proof: The first inequality is due to Conley et al.



Proof: The first inequality is due to Conley et al. For the second inequality, we plan to use the comparison with the harmonic mean c−1 and prove the following:



Proof: The first inequality is due to Conley et al. For the second inequality, we plan to use the comparison with the harmonic mean c−1 and prove the following: 2 2 2 2 ¯ H ≤ 5+ [1 + (1 + ε) ] c−1 ≤ 5 + [1 + (1 + ε) ] A. 1+ε 1+ε



A direct argument is used to show that if the principal curvatures are bounded away from 0, then c−1 ≤ A.



A direct argument is used to show that if the principal curvatures are bounded away from 0, then c−1 ≤ A. By using the notations |k1 | = a, |k2 | = b, |k3 | = c, the claim left to prove reduces to the following statement: if a, b, c ∈ [1, 1 + ε], then 1 1 1 2 (a + b + c) + + ≤5+ [1 + (1 + ε)2 ]. a b c 1+ε



A direct argument is used to show that if the principal curvatures are bounded away from 0, then c−1 ≤ A. By using the notations |k1 | = a, |k2 | = b, |k3 | = c, the claim left to prove reduces to the following statement: if a, b, c ∈ [1, 1 + ε], then 1 1 1 2 (a + b + c) + + ≤5+ [1 + (1 + ε)2 ]. a b c 1+ε By multiplying the terms in the left hand side of this expression, we obtain that we need to prove the following: 3+

a b c b c a 2 + + + + + ≤5+ [1 + (1 + ε)2 ]. b c b a a c 1+ε



Without any loss of generality, suppose that a ≥ b ≥ c. Then (a − b)(b − c) ≥ 0.



Without any loss of generality, suppose that a ≥ b ≥ c. Then (a − b)(b − c) ≥ 0. This yields ab + bc ≥ b 2 + ac. Dividing both sides by bc and by ab, respectively, we obtain first:



Without any loss of generality, suppose that a ≥ b ≥ c. Then (a − b)(b − c) ≥ 0. This yields ab + bc ≥ b 2 + ac. Dividing both sides by bc and by ab, respectively, we obtain first: a b a +1≥ + , c b c then



Without any loss of generality, suppose that a ≥ b ≥ c. Then (a − b)(b − c) ≥ 0. This yields ab + bc ≥ b 2 + ac. Dividing both sides by bc and by ab, respectively, we obtain first: a b a +1≥ + , c b c then 1+

c c b ≥ + . a b a



Without any loss of generality, suppose that a ≥ b ≥ c. Then (a − b)(b − c) ≥ 0. This yields ab + bc ≥ b 2 + ac. Dividing both sides by bc and by ab, respectively, we obtain first: a b a +1≥ + , c b c then 1+

c c b ≥ + . a b a

Adding term by term we get: a b c b a c + + + ≤2+ + . b c b a c a Thus it is always true that for a, b, c > 0, with a ≥ b ≥ c, by adding the same quantity both sides, we have a c b c a a b c + + + + + ≤2+2 + . b c b a a c c a Bogdan D. Suceav˘ a California State University, Fullerton

(1)


We estimate the right hand side term of this expression. Since 1 1 ≤ a ≤ 1 + ε, we have 1+ε ≤ 1a ≤ 1. From such estimates, by multiplying term by term double inequalities with strictly positive terms, we obtain



We estimate the right hand side term of this expression. Since 1 1 ≤ a ≤ 1 + ε, we have 1+ε ≤ 1a ≤ 1. From such estimates, by multiplying term by term double inequalities with strictly positive terms, we obtain a 1 ≤ ≤ 1 + ε. 1+ε c Denote by x = ca . Then [x − (1 + ε)] · x −

1 1+ε

≤ 0,

that is:



x+

1 1 ≤1+ε+ . x 1+ε

This estimate yields the best upper bound in (1):



x+

1 1 ≤1+ε+ . x 1+ε

This estimate yields the best upper bound in (1): a b c b c a 1 + + + + + ≤2+2 1+ε+ . b c b a a c 1+ε Adding 3 both sides yields



x+

1 1 ≤1+ε+ . x 1+ε

This estimate yields the best upper bound in (1): a b c b c a 1 + + + + + ≤2+2 1+ε+ . b c b a a c 1+ε Adding 3 both sides yields 3+

a b c b c a 2 + + + + + ≤5+ [1 + (1 + ε)2 ]. b c b a a c 1+ε



x+

1 1 ≤1+ε+ . x 1+ε

This estimate yields the best upper bound in (1): a b c b c a 1 + + + + + ≤2+2 1+ε+ . b c b a a c 1+ε Adding 3 both sides yields a b c b c a 2 + + + + + ≤5+ [1 + (1 + ε)2 ]. b c b a a c 1+ε which completes this proof. 3+



This result is from The amalgamatic curvature and the orthocurvatures of three dimensional hypersurfaces in the Euclidean space, Publicationes Mathematicae, Debrecen, Hungary 87 (2015), no. 1–2, 35–46.



THANK YOU !

