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typo in the TeX: complicated superscripts without enclosing braces
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joseph f. johnson
joseph f. johnson
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So we know that Hodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$$$⋆(dx^{i_1} \wedge ... \wedge dx^{i_p})= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where $p$ represents the $p$ in $p$-form and $n$ is the dimensional number.

My question is: How does hodge duality work on the imaginary number $i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

So we know that Hodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where $p$ represents the $p$ in $p$-form and $n$ is the dimensional number.

My question is: How does hodge duality work on the imaginary number $i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

So we know that Hodge duality works this way

$$⋆(dx^{i_1} \wedge ... \wedge dx^{i_p})= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where $p$ represents the $p$ in $p$-form and $n$ is the dimensional number.

My question is: How does hodge duality work on the imaginary number $i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

added 51 characters in body; edited tags
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Qmechanic ♦
Qmechanic ♦
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So we know that Hodge dualityHodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where p$p$ represents the p$p$ in p$p$-form and n$n$ is the dimensional number.

My question is: How does hodge duality work on the imaginary number "i"$i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

So we know that Hodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where p represents the p in p-form and n is the dimensional number.

My question is: How does hodge duality work on the imaginary number "i" on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

So we know that Hodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where $p$ represents the $p$ in $p$-form and $n$ is the dimensional number.

My question is: How does hodge duality work on the imaginary number $i$ on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.

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PhilosophicalPhysics
PhilosophicalPhysics
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Properties of Hodge Duality

So we know that Hodge duality works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where p represents the p in p-form and n is the dimensional number.

My question is: How does hodge duality work on the imaginary number "i" on one hand and on partial derivative like $\partial_x\alpha$ let us say on the other where $\alpha$ is a complex function.