An immersion has differential of rank , equivalently linearly independent . It has isothermal coordinates when its first fundamental form has , . A minimal surface has zero mean curvature.
For an isothermal immersion, differentiation of and gives
and similarly its scalar product with is zero. Its normal component is . The given mean curvature formula reduces to , so
Thus an isothermal immersion is minimal exactly when each coordinate is a harmonic function.
An explicit example is the Enneper surface, parametrized on by
Every coordinate has zero Laplacian. Direct differentiation gives and . Thus it is an isothermal immersion and is minimal everywhere.
It is not an open subset of any of the three listed surfaces. The parameter points and both map to , but their tangent planes differ: while . A plane, catenoid, or helicoid is an embedded smooth surface with a unique tangent plane at each point. If our whole image lay in any one of them, both rank- tangent planes at this point would have to equal its tangent plane, a contradiction.
The Euler totient function counts integers with . The Möbius function has , is zero when a prime square divides , and equals when is a product of distinct primes. Consequently
because for the sum over squarefree divisors is .
The Möbius inversion formula is . Substituting the divisor sum defining and collecting the coefficient of gives
This proof uses only finite sums, so no convergence hypothesis is needed.
Classifying by their greatest common divisor with gives : the class with greatest common divisor contains exactly integers. Applying the Möbius inversion formula yields
The Riemann zeta function is for . Absolute convergence justifies multiplication of Dirichlet series. The divisor identity for the Möbius function gives there. For both factors below converge absolutely, and their Dirichlet convolution is the Euler totient function:
A covering map is a continuous surjection for which each point has an open neighbourhood with a disjoint union of open sets, each mapped homeomorphically onto . A connected simply connected covering space is a universal cover.
There is a necessary connectedness qualification in the printed conclusion. Let be the path component of in . Then : one inclusion follows by projecting paths, and the other by lifting a path in from . Thus the asserted surjection onto all of requires to be path connected. Without it, take to be the identity of and to be two disjoint intervals; a selected component cannot cover both.
To prove the assertion about a component of a pulled-back universal cover, choose an evenly covered neighbourhood in . Local path connectedness of supplies a path-connected relatively open neighbourhood . Each inverse sheet over is path connected, so it lies wholly in or is disjoint from it. These sheets evenly cover under . Therefore is a covering map.
Under the classification of connected covering spaces, its subgroup consists of based loops in whose lifts close at . A loop viewed in closes in the universal cover exactly when its class in is trivial. Hence the subgroup is
Here , so the subgroup formula itself needs no change. Injectivity of the covering's induced fundamental-group map is the usual homotopy-lifting property.

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