According to the interesting answer of Rado, "trace" is an algebraic way to represent the dimension of fibres of a vector bundle on a compact Hausdorff space $X$. Every vector bundle on $X$ corresponds to an idempotent matrix valued function $A(x),\;x\in X$ . The dimension of fibre of a vector bundle is equal to $tr(A(x))$. Assuming $X$ is connected, this quantity is fix along $X$. I learned this from "Very basic noncommutative geometry" By Masoud Khalkhali (Can be found in arxiv)

According to the interesting answer of Rado, "trace" is an algebraic way to represent the dimension of fibres of a vector bundle on a compact Hausdorff space $X$. Every vector bundle on $X$ corresponds to an idempotent matrix valued function $A(x),\;x\in X$ . The dimension of fibre of a vector bundle is equal to $tr(A(x))$. Assuming $X$ is connected, this quantity is fix along $X$. I learned this from "Very basic noncommutative geometry" By Masoud Khalkhali (Can be found in arxiv).

The terminology "Trace" is also used in PDE as an operator which restricts functions in sobolov space $H^{s}(\Omega)$ to the boundary of $\Omega$

According to the interesting answer of Rado, trace is an algebraic way to represent the dimension of fibres of a vector bundle on a compact Hausdorff space $X$. Every vector bundle on $X$ corresponds to an idempotent matrix valued function $A(x),\;x\in X$ . The dimension of fibre of vector bundle is equal to $tr(A(x))$. I learned this from "Very basic noncommutative geometry" By Masoud Khalkhali (Can be find in arxiv)

According to the interesting answer of Rado, "trace" is an algebraic way to represent the dimension of fibres of a vector bundle on a compact Hausdorff space $X$. Every vector bundle on $X$ corresponds to an idempotent matrix valued function $A(x),\;x\in X$ . The dimension of fibre of a vector bundle is equal to $tr(A(x))$. Assuming $X$ is connected, this quantity is fix along $X$. I learned this from "Very basic noncommutative geometry" By Masoud Khalkhali (Can be found in arxiv)