几何学代写 Geometry代考2023

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几何学代写Geometry

点（几何学）Point (geometry)代写

• Line (geometry)线（几何学）
• Plane (geometry)平（几何学）

几何学的历史

The birth of geometry can be traced back to the time of the ancient Egyptians. Herodotus recounts that due to the erosion and deposition of the Nile floods, the extent of the Egyptians’ land changed from year to year, making it necessary to recalculate taxes. Therefore, it was necessary to invent land surveying techniques (geometry, in the original meaning of the term).

The development of practical geometry is very ancient, and since it could have many applications and was developed for those applications, in ancient times, practical geometry was sometimes reserved for a class of wise men with priestly attributes. In ancient Greece, mainly due to the influence of the Athenian philosopher Plato, and even before him, Anaximander of Miletus [citation needed], the use of rules and compasses became common (although these tools seem to have been invented elsewhere), and above all, new ideas using presentation techniques were born. Greek geometry became the basis for the development of geography, astronomy, optics, mechanics and other sciences, as well as various technologies (such as navigation technology).

In Greek civilization, in addition to Euclidean geometry and conic theory, which were still studied in schools, spherical geometry and trigonometry (plane and sphere) were born.

几何学相关课后作业代写

Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.

To show that $S^4$ has no symplectic structure, we will use the following fact from symplectic geometry: a compact symplectic manifold has even dimension.

Suppose that $S^4$ has a symplectic structure. Then $S^4$ is a compact symplectic manifold, so its dimension must be even. However, the dimension of $S^4$ is $4$, which is not even. Therefore, $S^4$ cannot have a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we will use the following fact: the product of two symplectic manifolds is symplectic if and only if both factors have even dimension.

Suppose that $S^2 \times S^4$ has a symplectic structure. Then both $S^2$ and $S^4$ are symplectic manifolds, so their dimensions must both be even. However, the dimension of $S^2$ is $2$, which is not even. Therefore, $S^2 \times S^4$ cannot have a symplectic structure.