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Date: 29-7-2021
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Date: 26-6-2021
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Date: 30-5-2021
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de Rham cohomology is a formal set-up for the analytic problem: If you have a differential k-form on a manifold
, is it the exterior derivative of another differential k-form
? Formally, if
then
. This is more commonly stated as
, meaning that if
is to be the exterior derivative of a differential k-form, a necessary condition that
must satisfy is that its exterior derivative is zero.
de Rham cohomology gives a formalism that aims to answer the question, "Are all differential -forms on a manifold with zero exterior derivative the exterior derivatives of
-forms?" In particular, the
th de Rham cohomology vector space is defined to be the space of all
-forms with exterior derivative 0, modulo the space of all boundaries of
-forms. This is the trivial vector space iff the answer to our question is yes.
The fundamental result about de Rham cohomology is that it is a topological invariant of the manifold, namely: the th de Rham cohomology vector space of a manifold
is canonically isomorphic to the Alexander-Spanier cohomology vector space
(also called cohomology with compact support). In the case that
is compact, Alexander-Spanier cohomology is exactly singular cohomology.
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دخلت غرفة فنسيت ماذا تريد من داخلها.. خبير يفسر الحالة
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ثورة طبية.. ابتكار أصغر جهاز لتنظيم ضربات القلب في العالم
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قسم شؤون المعارف ووفد من جامعة البصرة يبحثان سبل تعزيز التعاون المشترك
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