nyuuzyou/wikis huggingface dataset:
Conditions for Homogeneity/Straight Line

== Theorem ==
The line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ is homogeneous $\beta = 0$.
== Proof ==
=== Sufficient Condition ===
Let the line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ be homogeneous.
Then the origin $\tuple {0, 0}$ lies on $L$.
Hence:
=== Necessary Condition ===
Let the equation of $L$ be $\alpha_1 x_1 + \alpha_2 x_2 = 0$.
Then:
:$0 = \alpha_1 0 + \alpha_2 0 \in L$
and so:
:$\tuple {0, 0}$ lies on $L$
Hence $L$ is homogeneous by definition.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Algebra
Category:Analytic Geometry
Conditions for Integer to have Primitive Root

== Theorem ==
Let $n \in \Z: n > 1$.
Then $n$ has a primitive root one of the following holds:
:$n = 2$
:$n = 4$
:$n = p^k$
:$n = 2 p^k$
where $p > 2$ is prime and $k \ge 1$.
== Proof of Necessity ==
This proof comes in several sections, so as to be able to cover all cases.
=== The cases where $n = 2$ and $n = 4$ ===
$1$ is trivially a primitive root of $2$.
$3$ is a primitive root of $4$, as:
:$\map \phi 4 = 2$
and:
:$3^2 = 9 \equiv 1 \pmod 4$
=== Power of $2$ greater than $4$: No Primitive Root ===
Let $n = 2^k$ where $k \ge 3$.
From Euler Phi Function of Prime Power: Corollary:
:$\map \phi {2^k} = 2^{k - 1}$
The $2^{k - 1}$ least positive residues prime to $2^k$ are the odd integers up to $2^k - 1$.
It is to be shown that for any odd integer $a$ there exists a positive integer $r < 2^{k-1}$ such that $a \equiv 1 \pmod {2^k}$.
We assert that $r = 2^{k - 2}$ has exactly this property, which we prove by induction.
For all $k \in \N_{>0}$, let $\map P k$ be the proposition:
:$a^{2^{k - 2} } \equiv 1 \pmod {2^k}$
==== Basis for the Induction ====
$\map P 3$ is true, as follows:
:$k = 3 \implies 2^k = 8, 2^{k - 2} = 2^1 = 2$
The odd integers less than $8$ are $1, 3, 5, 7$.
So:
Thus $\map P 3$ holds.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P h$ is true, where $h \ge 3$, then it logically follows that $\map P {h + 1}$ is true.
So this is our induction hypothesis:
:$a^{2^{h - 2} } \equiv 1 \pmod {2^h}$
Then we need to show:
:$a^{2^{h - 1} } \equiv 1 \pmod {2^{h + 1} }$
==== Induction Step ====
We can write our induction hypothesis more conveniently as:
:$\exists m \in \Z: a^{2^{h - 2} } = 1 + m 2^h$
Hence:
But:
So $\map P h \implies \map P {h + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall k \ge 3: a^{2^{k - 2} } \equiv 1 \pmod {2^k}$
=== Two Coprime Divisors Greater than 2: No Primitive Root ===
Let us take $n = r s$ where $r > 2, s > 2, r \perp s$.
Let $a \perp r s$.
From Euler Phi Function is Even for Argument greater than 2, $\map \phi r$ and $\map \phi s$ are both even.
So $\dfrac {\map \phi r} 2$ and $\dfrac {\map \phi s} 2$ are both integers.
Hence $\dfrac {\map \phi {r s} } 2$ is an integer.
As $a \perp r s$ we have that $a \perp r$
So from Euler's Theorem (Number Theory):
:$a^{\map \phi r} \equiv 1 \pmod r$
So putting $k = \dfrac {\map \phi {r s} } 2$:
Interchanging the roles of $r$ and $s$ shows also that:
:$a^k \equiv 1 \pmod s$
So we see that:
:$a^k \equiv 1 \pmod {r s}$
Thus for any $a$, we have:
:$a^k \equiv 1 \pmod n$
where:
:$k = \dfrac {\map \phi n} 2 < \map \phi n$
Hence $n$ has no primitive root.
Hence we see that in order for $n$ to have a primitive root, it is necessary that one of the following hold:
:$n = 2$
:$n = 4$
:$n = p^k$
:$n = 2 p^k$
where $p > 2$ is prime and $k \ge 1$.
== Proof of Sufficiency ==
It remains to cover the cases where $n = 2$ and $n = 4$.
This it is to be shown that if either of the following hold:
:$n = p^k$
:$n = 2 p^k$
where $p > 2$ is prime and $k \ge 1$, then $n$ has a primitive root.
== Historical Note ==
Category:Number Theory
Conditions for Integral Functionals to have same Euler's Equations

== Theorem ==
Let $\mathbf y$ be a real $n$-dimensional vector-valued function.
Let $\map F {x, \mathbf y, \mathbf y'}$, $\map \Phi {x, \mathbf y}$ be real functions.
Let $\Phi$ be twice differentiable.
Let:
Let $J_1$, $J_2$ be functionals such that:
:$\ds J_1 \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
:$\ds J_2 \sqbrk {\mathbf y} = \int_a^b \paren {\map F {x, \mathbf y, \mathbf y'} + \map \Psi {x, \mathbf y, \mathbf y'} } \rd x$ 
Then $J_1$ and $J_2$ have same Euler's Equations.
== Proof ==
According to Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions:
Euler's Equations for functional $J_1$ are: 
:$\ds F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} = 0$
Equivalently, for $J_2$ we have
Furthermore:
Since $\Phi$ is twice differentiable, by Schwarz-Clairaut Theorem partial derivatives commute and:
:$\Psi_{\mathbf y} - \dfrac \d {\d x} \Psi_{\mathbf y'} = 0$
Therefore, $J_1$ and $J_2$ have same Euler's Equations.
== Sources ==
* : $\S 2.9$: The Fixed End Point Problem for n Unknown Functions
Category:Calculus of Variations
Conditions for Internal Ring Direct Sum

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\sequence {\struct {S_k, +, \circ} }$ be a sequence of subrings of $R$.
Then $R$ is the ring direct sum of $\sequence {S_k}_{1 \mathop \le k \mathop \le n}$ :
:$(1): \quad R = S_1 + S_2 + \cdots + S_n$
:$(2): \quad \sequence {\struct {S_k, +} }_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups of $\struct {R, +}$
:$(3): \quad \forall k \in \closedint 1 n: S_k$ is an ideal of $R$.
== Proof ==
Let $S$ be the cartesian product of $\sequence {\struct {S_k, +} }$
First note that $\phi$ is a group homomorphism from $\struct {S, +}$ to $\struct {R, +}$, as:
:$\ds \sum_{j \mathop = 1}^n \paren {x_j + y_j} = \sum_{j \mathop = 1}^n x_j + \sum_{j \mathop = 1}^n y_j$ from Associativity on Indexing Set.
So $\phi$ is a ring homomorphism :
:$\ds \paren {\sum_{j \mathop = 1}^n x_j} \circ \paren {\sum_{j \mathop = 1}^n y_j} = \sum_{j \mathop = 1}^n \paren {x_j \circ y_j}$
Now, let:
:$\struct {S, +, \circ}$ be the cartesian product of $\sequence {\struct {S_k, +, \circ} }$
:$\phi: S \to R$ be the mapping defined as:
::$\map \phi {x_1, x_2, \ldots, x_n} = x_1 + x_2 + \cdots x_n$
Clearly $\phi$ is a surjection $(1)$ holds.
By Internal Direct Product Generated by Subgroups, $\phi$ is a ring isomorphism $(1)$ and $(2)$ hold.
Let $\phi: S \to R$ be a ring isomorphism.
By Canonical Injection from Ideal of External Direct Sum of Rings, $\map {\inj_k} {S_k}$ is an ideal of $S$.
So $\map \phi {\map {\inj_k} {S_k} }$ is an ideal of $R$.
But $\phi$ and $\pr_k$ coincide on $\map {\inj_k} {S_k}$.
So:
:$\map \phi {\map {\inj_k} {S_k} } = \map {\pr_k} {\map {\inj_k} {S_k} } = S_k$
and so $(3)$ holds.
Now suppose $(3)$ holds, and the $S_k$ are all ideals of $R$.
By $(2)$ and by definition of independent subgroups:
:$i \ne j \implies S_i \cap S_j = \set {0_R}$
So for all $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in S$:
because as $S_i, S_j$ are ideals, we have:
:$x_i \circ j_j = S_i \cap S_j = \set {0_R}$
Hence the three conditions are sufficient for $\phi$ to be a ring isomorphism.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.9$
* : $\S 3.1$: Direct sums: Lemma $3.3$
Category:Subrings
Category:Ideal Theory
Category:Direct Sums of Rings
Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Corollary

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Let $\struct {S_2, \preccurlyeq_2}$ have neither a greatest element nor a smallest element.
Then:
:$\preccurlyeq_l$ is a lattice ordering
:
:$\preccurlyeq_1$ is a total ordering
and:
:$\preccurlyeq_2$ is a lattice ordering.
== Proof ==
=== Sufficient Condition ===
Let $\preccurlyeq_l$ be a lattice ordering.
We are given that $\struct {S_2, \preccurlyeq_2}$ has neither a greatest element nor a smallest element.
From Condition $(2)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice, it follows that $\preccurlyeq_1$ is a total ordering.
Because $\struct {S_2, \preccurlyeq_2}$ has no smallest element, it follows from Condition $(4)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice that every doubleton subset of $S_2$ admits a supremum.
Because $\struct {S_2, \preccurlyeq_2}$ has no greatest element, it follows from Condition $(5)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice that every doubleton subset of $S_2$ admits an infimum.
Thus every doubleton subset of $S_2$ admits a supremum and an infimum.
That is, $\preccurlyeq_2$ is a lattice ordering.
=== Necessary Condition ===
Let:
:$\preccurlyeq_1$ be a total ordering
and:
:$\preccurlyeq_2$ be a lattice ordering.
From Totally Ordered Set is Lattice, $\preccurlyeq_1$ is a lattice ordering.
Thus Condition $(1)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice holds.
We are given that $\preccurlyeq_1$ be a total ordering.
Thus Condition $(2)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice holds.
We are given that $\preccurlyeq_2$ is a lattice ordering.
Hence every doubleton subset of $S_2$ admits a supremum and an infimum.
Thus Conditions $(3)$, $(4)$ and $(5)$ of Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice hold.
Hence from Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice:
:$\preccurlyeq_l$ is a lattice ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.19 \ \text{(c)}$
Category:Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice
Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 1

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Let $\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$.
Then:
:$x_1 \preccurlyeq_1 y_1$
== Proof ==
Recall the definition of lexicographic order:
Then:
Category:Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice
Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice/Lemma 2

== Theorem ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\preccurlyeq_l$ denote the lexicographic order on $S_1 \times S_2$:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} \iff \tuple {x_1 \prec_1 y_1} \lor \tuple {x_1 = y_1 \land x_2 \preccurlyeq_2 y_2}$
Let $x_1$ and $y_1$ be non-comparable elements in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
Then $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ are non-comparable elements in $S_1 \times S_2$:
:$\lnot \paren {\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2} }$ and $\lnot \paren {\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2} }$
== Proof ==
Suppose $x_1$ and $y_1$ are non-comparable elements in $S_1$:
:$\lnot \paren {x_1 \preccurlyeq_1 y_1}$ and $\lnot \paren {y_1 \preccurlyeq_1 x_1}$
 that:
:$\tuple {x_1, x_2} \preccurlyeq_l \tuple {y_1, y_2}$ or $\tuple {y_1, y_2} \preccurlyeq_l \tuple {x_1, x_2}$
Then from Lemma $1$:
:$x_1 \preccurlyeq_1 y_1$ or $y_1 \preccurlyeq_1 x_1$
This contradicts our assumption.
Hence the result by Proof by Contradiction.
Category:Conditions for Lexicographic Order on Pair of Ordered Sets to be Lattice
Conditions for Limit Function to be Limit Minimizing Function of Functional

== Theorem ==
Let $y$ be a real function.
Let $J \sqbrk y$ be a functional.
Let $\sequence {y_n}$ be a minimizing sequence of $J$.
Let:
:$\ds \lim_{n \mathop \to \infty} y_n = \hat y$
Suppose $J$ is lower semicontinuous at $y = \hat y$.
Then:
:$\ds J \sqbrk {\hat y} = \lim_{n \mathop \to \infty} J \sqbrk {y_n}$
== Proof ==
By definition of minimizing sequence:
:$\ds \inf_y J \sqbrk y = \lim_{n \mathop \to \infty} J \sqbrk {y_n}$
Any mapping from this sequence either minimises the functional or not.
This is true for the limit mapping as well:
:$\ds J \sqbrk {\hat y} \ge \inf_y J \sqbrk y$
By assumption, $J$ is lower semicontinuous at $\hat y$:
:$\ds \forall \epsilon_1 > 0: \forall n \in \N: \exists N \in \N: \paren {n > N} \implies \paren {J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon_1}$
Hence, for sufficiently large $n$:
:$\ds \inf_y J \sqbrk y \le J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$
By the definition of minimizing sequence, where the label $n$ has been replaced with $m$:
:$\ds \forall \epsilon_2: \forall m \in \N: \exists M \in \N: \paren {m > M} \implies \paren {\size {J \sqbrk {y_m} - \inf_y J \sqbrk y} < \epsilon_2}$
From this, for sufficiently large $m$:
:$\ds J \sqbrk {y_m} - \epsilon_2 < \inf_y J \sqbrk y$
 
Then:
:$J \sqbrk {y_m} - \epsilon_2 < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon_1$
Here $\epsilon_1$, $n$ have similar properties like $\epsilon_2$, $m$, but are otherwise arbitrary and independent.
Let $\epsilon_1 = \epsilon_2 = \epsilon$, $n = m$, $M = N$.
Arbitrariness is still not affected. 
Then:
:$J \sqbrk {y_n} - \epsilon < J \sqbrk {\hat y} < J \sqbrk {y_n} + \epsilon$
Subtract $J \sqbrk {y_n}$ from all the terms.
Hence:
:$-\epsilon < J \sqbrk {\hat y} - J \sqbrk {y_n} < \epsilon$
or:
:$\size {J \sqbrk {y_n} - J \sqbrk {\hat y} } < \epsilon$
Therefore this relation inherits all the constraints on its values $n$, $N$, $\epsilon$, and by definition is a limit:
:$\ds \lim_{n \mathop \to \infty} J \sqbrk {y_n} = J \sqbrk {\hat y}$
== Sources ==
* : $\S 8.39$: Minimizing Sequences
Category:Calculus of Variations
Conditions for Ordering in Ordered Group to be Directed

== Theorem ==
Let $\struct {G, \odot, \preccurlyeq}$ be an ordered group whose identity element is $e$.
Then:
:$\preccurlyeq$ is a directed ordering
:for every $x \in G$ there exist $y, z \in G$ such that $e \preccurlyeq y$, $e \preccurlyeq z$ and $x = y \odot z^{-1}$.
== Proof ==
=== Sufficient Condition ===
Let $\preccurlyeq$ be a directed ordering.
By definition of directed ordering:
:$\forall x, z \in G: \exists y \in G: x \preccurlyeq y$ and $z \preccurlyeq y$
Let $x \in G$ be arbitrary.
Let $e = \paren {y \odot x^{-1} }^{-1} \odot z$.
Then:
Then from:
we get:
and we see that:
:$\exists y, z \in G: e \preccurlyeq y, e \preccurlyeq z, x = y \odot z^{-1}$
=== Necessary Condition ===
Let $\preccurlyeq$ be such that:
:for every $x \in G$ there exist $y, z \in G$ such that $e \preccurlyeq y$, $e \preccurlyeq z$ and $x = y \odot z^{-1}$.
Let $x \in G$ be arbitrary.
By the hypothesis, there exist $z, w \in G$ such that $e \preccurlyeq z$, $e \preccurlyeq w$ and $x = z \odot w^{-1}$.
We have:
Similarly, for another arbitrary $y \in G$, there exists some $g \in G$ such that $e \preccurlyeq g$ and $y \preccurlyeq g$.
Then we have:
This shows that $z \odot g$ is an upper bound of $\set {x, y}$.
By , $z \odot g \in G$.
Hence $\preccurlyeq$ is a directed ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.13$
Category:Ordered Groups
Category:Directed Orderings
Conditions for Preservation of Covergence in Test Function Space under Differentiation

== Theorem ==
For all $n \in \N$ let $\Phi_n, \phi_n \in \map \DD \R$ be test functions.
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Let $\phi_n$ be such that:
:$\ds \int_{-\infty}^\infty \map {\phi_n} x \rd x = 0$
Let $\Phi_n$ be such that $\Phi_n' = \phi_n$.
Let $\sequence {\Phi_n}_{n \mathop \in \N}$ and $\sequence {\phi_n}_{n \mathop \in \N}$ be sequences in $\map \DD \R$.
Suppose $\sequence {\phi_n}_{n \mathop \in \N}$ converges to $\mathbf 0$ in $\map \DD \R$.
Then $\sequence {\Phi_n}_{n \mathop \in \N}$ converges to $\mathbf 0$ in $\map \DD \R$ as well.
== Proof ==
By Characterization of Derivative of Test Function we have that for every $\phi_n$ there is a unique $\Phi_n$ such that:
:$\ds \map {\Phi_n} x = \int_{-\infty}^x \map {\phi_n} x \rd x$
Let $K = \closedint {-a} a$ be a closed real interval.
Suppose that $\sequence {\phi_n}_{n \mathop \in \N}$ is supported on $K$.
Hence:
:$\forall n \in \N : \forall x \in \R \setminus K : \map {\phi_n} x = 0$
From assumption about $\phi_n$ it follows that:
:$\forall n \in \N : \forall x \in \R \setminus K : \map {\Phi_n} x = 0$
Furthermore, for all $x \in \R$ we have that:
By assumption, $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.
Hence, $\sequence {\phi_n}$ converges uniformly to $\mathbf 0$ on $\R$:
:$\ds \forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in \R} \size {\map {\phi_n} x} < \epsilon$
Therefore:
 
Thus, $\sequence {\map {\Phi_n} x}$ converges uniformly on $\R$ to $\mathbf 0$.
Since $\Phi'_n = \phi_n$, it follows that $\Phi_n^{\paren k} = \phi_n^{\paren {k - 1} }$ with $k \in \N_{>0}$.
Hence, for all $k \in \N_{> 0}$ we have that:
 
By assumption, $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.
Hence, for all $k \in \N_{>0}$ we have that $\sequence {\phi_n^{\paren {k - 1} } }$ converges uniformly to $\mathbf 0$ on $\R$:
:$\ds \forall k \in \N_{>0} : \forall \epsilon \in \R_{>0}: \exists N \in \R: \forall n \ge N: \sup_{x \mathop \in \R} \size {\map {\phi_n^{\paren k} } x} < \epsilon$
Then:
Hence, for all $k \in \N_{>0}$ we have that $\sequence {\Phi_n^{\paren k} }$ converges uniformly to $\mathbf 0$ on $\R$.
By definition, $\sequence {\Phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Test Functions
Category:Convergence
Conditions for Pullback of Pseudo-Riemannian Metric to be Nondegenerate

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a pseudo-Riemannian manifold of signature $\tuple {r, s}$.
Let $M$ be a smooth hypersurface in $\tilde M$.
Let $\iota : M \to \tilde M$ be the inclusion mapping.
Let $N_p M$ be the normal space at $p \in M$.
Then the pullback tensor field $\iota^* \tilde g$ is nondegenerate 
:$\forall p \in M : \forall v \in N_p M : v \ne 0 : \map {\tilde g} {v, v} \ne 0$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Pseudo-Riemannian Metrics
Category:Definitions/Pseudo-Riemannian Geometry
Conditions for Quotient Map from Riemannian Manifold to its Quotient by Discrete Lie Group to be Normal Riemannian Covering

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a Riemannian manifold.
Let $\Gamma$ be a discrete Lie group.
Let $\pi : \tilde M \to \tilde M ~/~ \Gamma$ be a quotient map.
Suppose the action of $\Gamma$ is smooth, free, proper, and isometric.
Then $\tilde M ~/~ \Gamma$ has a unique Riemannian metric $g$ such that $\pi$ is a normal Riemannian covering.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Conditions for Smooth Normal Covering Map to be Riemannian Covering

== Theorem ==
Let $\struct {\tilde M, \tilde g}$, $\struct {M, g}$ be Riemannian manifolds.
Let $\pi : \tilde M \to M$ be a smooth normal covering map.
Let $\tilde g$ be invariant under all covering automorphisms.
Then there exists a unique $g$ such that $\pi$ is a Riemannian covering.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Conditions for Strong Minimum of Functional

== Theorem ==
Let $\mathbf y$ be an $n$-dimensional vector such that:
:$\map {\mathbf y} a = A$ and $\map {\mathbf y} b = B$
Let $J$ be a functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let $\gamma$ be an extremal curve of $J$.
Let the following be the field of the functional $J$:
:$\mathbf y' = \map {\boldsymbol \psi} {x, \mathbf y}$
Let $R$ be an open region containing $\gamma$ and have the field $\boldsymbol \psi$ defined as $\forall \paren {x, \mathbf y} \in R$.
Let $\mathbf w$ be a finite vector.
Suppose that:
:$\forall \paren {x, \mathbf y} \in R: \map E {x, \mathbf y, \boldsymbol \psi, \mathbf w}\ge 0$
where $E$ is Weierstrass E-Function.
Then $J$ has a strong minimum for $\gamma$.
== Proof ==
By definition, the increment of $J$ is:
:$\ds \Delta J = \int_{\gamma^*} \map F {x, \mathbf y, \mathbf y'} \rd x - \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$
where $\gamma$ and $\gamma^*$ are curves described by $\paren {x, \map {\mathbf y} x}$ and $\paren {x, \paren {\mathbf y^*} x}$ respectively, such that:
:$\map {\mathbf y^*} x - \map {\mathbf y} x = \map {\mathbf h} x$
Consider Hilbert's invariant integral:
Since the integrand is full differential, the integral does not depend on the shape of $ \Gamma $, but only on its endpoints.
Therefore, for $\Gamma = \gamma$ and $\Gamma = \gamma^*$ the value of the integral is the same.
Since $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$ determines boundary conditions for $\gamma$, $\Gamma = \gamma$ is one of the trajectories of the field $\map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$.
Hence, $\d \mathbf y$ is constrained by $\d \mathbf y = \boldsymbol \psi \rd x$:
:$\ds \map g {x, \mathbf y} = \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x$
Thus, $\map g {x, \mathbf y'}$ can be written in two different ways.
:$\ds \int_\gamma \map F {x, \mathbf y, \mathbf y'} \rd x = \int_{\gamma^*} \paren {\paren {\map F {x, \mathbf y, \boldsymbol \psi} - \boldsymbol \psi F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} } \rd x + F_{\mathbf y'} \paren {x, \mathbf y, \boldsymbol \psi} \rd \mathbf y}$
Substitute this into the expression for $\Delta J$:
By assumption:
:$\map E {x, \mathbf y, \boldsymbol \psi, \mathbf y'} \ge 0$
Hence the integrand is bounded below by $0$ and above by its maximum in the interval of integration.
Hence, the integral is bounded below by $0$ and above by some positive number.
Therefore:
:$\Delta J \ge 0$
== Sources ==
* : $\S 6.34$: The Weierstrass E-Function. Sufficient Conditions for a Strong Extremum
Category:Calculus of Variations
Conditions for Subjective Smooth Submersion from Riemannian Manifold to its Orbit Space to be Riemannian Submersion

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be a Riemannian manifold.
Let $\pi : \tilde M \to M$ be a surjective smooth submersion.
Let $G$ be a Lie group acting on $\tilde M$.
Suppose the action of $G$ is smooth, free, proper, and isometric.
Let $M = \tilde M \setminus G$ be the orbit space.
Then $M$ has a unique smooth manifold structure and Riemannian metric $g$ such that $\pi$ is a Riemannian submersion.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Conditions for Subjective Smooth Submersion to be Riemannian Submersion

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ and $\struct {M, g}$ be Riemannian manifolds.
Let $\pi : \tilde M \to M$ be a surjective smooth submersion.
Let $G$ be a group acting on $\tilde M$.
Suppose the action of $G$ is isometric, vertical, and transitive on fibers.
Then there exists a unique Riemannian metric $g$ such that $\pi$ is a Riemannian submersion.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry