How to visualize 2-form or exterior product?

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In summary, we can visualize 1-form by contour lines, since a 1-form / gradient sort of represents how fast the function changes. I wonder whether we can visualize 2-form df ^ dg by intersection of two sets of contour lines for f and g, or maybe something of a similar nature?
  • #1
lriuui0x0
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We can visualize 1-form by contour lines, since a 1-form / gradient sort of represents how fast the function changes. I wonder whether we can visualize 2-form df ^ dg by intersection of two sets of contour lines for f and g, or maybe something of a similar nature?
 
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  • #2
lriuui0x0 said:
We can visualize 1-form by contour lines, since a 1-form / gradient sort of represents how fast the function changes. I wonder whether we can visualize 2-form df ^ dg by intersection of two sets of contour lines for f and g, or maybe something of a similar nature?
It's rather the (oriented) area defined by df and dg.
 
  • #3
fresh_42 said:
It's rather the (oriented) area defined by df and dg.
Let's say we have a grid of contour lines defined by df and dg, consecutive contour lines have their function values changed by one. Then does each grid cell correspond to area one?
 
  • #4
lriuui0x0 said:
Let's say we have a grid of contour lines defined by df and dg, consecutive contour lines have their function values changed by one. Then does each grid cell correspond to area one?
Such visualizations have their limits. It is better in this context to consider ##df## and ##dg## as coordinate vectors rather than infinitesimals.

##a_1 \wedge a_2 \wedge \ldots \wedge a_n## is the oriented volume spanned by the vectors ##a_k.## We have ##n=2## here, and ##a_k## are differential ##1-##forms, covectors.

Of course, you can always consider differential forms as smooth sections of cotangent bundles. This is per se a geometric definition, even if a more sophisticated one, but nevertheless, geometric.

Here is a more algebraic description:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/
 
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