"exterior derivative" meaning in English

See exterior derivative in All languages combined, or Wiktionary

Noun

Audio: LL-Q1860 (eng)-Flame, not lame-exterior derivative.wav Forms: exterior derivatives [plural]
Head templates: {{en-noun}} exterior derivative (plural exterior derivatives)
  1. (calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0. Wikipedia link: exterior derivative Categories (topical): Calculus
    Sense id: en-exterior_derivative-en-noun-D-oZNg1F Categories (other): English entries with incorrect language header, Pages with 1 entry, Pages with entries Topics: calculus, mathematics, sciences

Inflected forms

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          "text": "The exterior derivative of a “scalar”, i.e., a function f=f(x¹,x²,...,xⁿ) where the xⁱ’s are coordinates of ℝⁿ, is df=∂f/∂x¹dx¹+∂f/∂x²dx²+...+∂f/∂xⁿdxⁿ."
        },
        {
          "text": "The exterior derivative of a k-blade f,dx∧dx∧...∧dx is df∧dx∧dx∧...∧dx."
        },
        {
          "text": "The exterior derivative d may be though of as a differential operator del wedge: ∇∧, where ∇=∂/∂x¹dx¹+∂/∂x_2dx²+...+∂/∂xⁿdxⁿ. Then the square of the exterior derivative is d²=∇∧∇∧=(∇∧∇)∧=0∧=0 because the wedge product is alternating. (If u is a blade and f a scalar (function), then fu≡f∧u, so d(fu)=∇∧(fu)=∇∧(f∧u)=(∇∧f)∧u=df∧u.) Another way to show that d²=0 is that partial derivatives commute and wedge products of 1-forms anti-commute (so when d² is applied to a blade then the distributed parts end up canceling to zero.)"
        }
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        "A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
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        "(calculus) A differential operator which acts on a differential k-form to yield a differential (k+1)-form, unless the k-form is a pseudoscalar, in which case it yields 0."
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          "text": "The exterior derivative of a “scalar”, i.e., a function f=f(x¹,x²,...,xⁿ) where the xⁱ’s are coordinates of ℝⁿ, is df=∂f/∂x¹dx¹+∂f/∂x²dx²+...+∂f/∂xⁿdxⁿ."
        },
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          "text": "The exterior derivative of a k-blade f,dx∧dx∧...∧dx is df∧dx∧dx∧...∧dx."
        },
        {
          "text": "The exterior derivative d may be though of as a differential operator del wedge: ∇∧, where ∇=∂/∂x¹dx¹+∂/∂x_2dx²+...+∂/∂xⁿdxⁿ. Then the square of the exterior derivative is d²=∇∧∇∧=(∇∧∇)∧=0∧=0 because the wedge product is alternating. (If u is a blade and f a scalar (function), then fu≡f∧u, so d(fu)=∇∧(fu)=∇∧(f∧u)=(∇∧f)∧u=df∧u.) Another way to show that d²=0 is that partial derivatives commute and wedge products of 1-forms anti-commute (so when d² is applied to a blade then the distributed parts end up canceling to zero.)"
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This page is a part of the kaikki.org machine-readable English dictionary. This dictionary is based on structured data extracted on 2024-12-21 from the enwiktionary dump dated 2024-12-04 using wiktextract (d8cb2f3 and 4e554ae). The data shown on this site has been post-processed and various details (e.g., extra categories) removed, some information disambiguated, and additional data merged from other sources. See the raw data download page for the unprocessed wiktextract data.

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