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Let $C$ \mathcal{C}$be a cocomplete$R$-linear tensor category. Many notions in of commutative algebra may be internalized to$C$. \mathcal{C}$. For example an algebra is an object $A$ in $C$ \mathcal{C}$together with morphisms$e : 1 \to A$(unit) and$m: A \otimes A \to A$(multiplication) satisfying the usual laws. The$n$th n$-th symmetric power $\text{Sym}^n(X)$ of an object $X$ is the quotient of $X^{\otimes n}$ by $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of two maps the $(X^{\otimes n})^{n!n!$ symmetries $X^{\otimes n} \to X^{\otimes n}$. Then $\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$ is an algebra object and $\text{Sym}$ is in fact left adjoint to the forgetful functor from algebras in $C$ \mathsf{CAlg}(\mathcal{C}) \to $C$.\mathsf{CAlg}(\mathcal{C})$. But now what about the exterior power$\wedge^n(X)$? \Lambda^n(X)$? It is clear how to define $X^{\otimes n}$ modulo $x_1 \otimes ... \otimes x_n = \text{sign}(\sigma) text{sgn}(\sigma) \cdot x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$ in this context, which one might call the skew-symmetric anti-symmetric power. $\mathrm{ASym}^n(X)$. The correct definition of the exterior power also has to mod out $... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to $C$. \mathcal{C}$, even for$n=2$. The reason is that there is no morphism$X \to X \otimes X$which acts like$a \mapsto a \otimes a$. Another idea would be to define$\wedge(X)$\Lambda(X)$ as a graded-commutative algebra object with the usual universal property, classifying maps from morphisms $f$ on $X$ which satisfy something like $f(x)^2=0$, but again it is unclear how to formulate this in $C$.\mathcal{C}$. If this is not possible at all, which additional structure on$C$\mathcal{C}$ do we need in order to define exterior powers within them? Is this some categorified $\lambda$-ring structure? This structure should be there in the case of usual module categories .

PS: I am interested in exterior powers since they should reduce locally free objects to line objects via the Plücker embedding(over rings or even ringed spaces). The $d$th anti-symmetric power of $1^n$ Of course there is not no problem when $1^{\binom{n}{d}} = 2 \wedge^d 1^n$ in generalR^*$, therefore it is not because then the correct notion hereexterior power equals the anti-symmetric power. EDIT: The question was also discussed in a blog post. Here is a more specific (and a bit stronger) formulation: Is there an some$R[S_n]$-module R[\Sigma_n]$-module $T$, such that for every $R$-module $X$, we have that $T \otimes_{R[S_n]} otimes_{R[\Sigma_n]} X^{\otimes n}$ equals $\wedge^n n} \cong \Lambda^n X := X^{\otimes n}/(... \otimes x ... \otimes x ...)$? Probably not, because $T$ should be a quotient of $R[S_n]$, and the corresponding ideal cannot generate squares.

EDIT:

Concerning the "hidden extra structure" in the case of modules, I have made the following observation: Let the ground base ring be $\mathbb{Z}$ (\mathbb{Z}$, or more generally a ring$R$in which$r^2 - r \in 2R$for all$r \in R$; this indluces includes boolean rings such as$\mathbb{F}_2$and also$\mathbb{Z}/n$). Then \mathbb{Z}/n$. If $X$ is an $R$-module, then there is a well-defined(!) map homomorphism $f_n : X^{\otimes nX^{\otimes~ n-1} \to \text{Asym}(X), text{ASym}^n(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$in the case of modules. The , and its cokernel of $\oplus_{n \geq 2} f_n$ is $\wedge(X)$.

EDIT: If there won't be any definition in the general case, one might define $\lambda$ tensor categories as categorified $\lambda$-rings. But I'm still curious why have quite often a distinguished $\lambda$ structure, coming from a specific quotient of the tensor power ...\Lambda^n(X)$. 6 added 277 characters in body Let$C$be a cocomplete$R$-linear tensor category. Many notions in commutative algebra may be internalized to$C$. For example an algebra is an object$A$in$C$together with morphisms$e : 1 \to A$(unit) and$m: A \otimes A \to A$(multiplication) satisfying the usual laws. The$n$th symmetric power$\text{Sym}^n(X)$of an object$X$is the quotient of$X^{\otimes n}$by$x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of two maps$(X^{\otimes n})^{n!} \to X^{\otimes n}$. Then$\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$is an algebra object and$\text{Sym}$is in fact left adjoint to the forgetful functor from algebras in$C$to$C$. But now what about the exterior power$\wedge^n(X)$? It is clear how to define$X^{\otimes n}$modulo$x_1 \otimes ... \otimes x_n = \text{sign}(\sigma) x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$in this context, which one might call the skew-symmetric power. The correct definition of the exterior power also has to mod out$... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to$C$. The reason is that there is no morphism$X \to X \otimes X$which acts like$a \mapsto a \otimes a$. Another idea would be to define$\wedge(X)$as a graded-commutative algebra object with the usual universal property, classifying maps from$X$which satisfy something like$f(x)^2=0$, but again it is unclear how to formulate this in$C$. If this is not possible at all, which additional structure on$C$do we need in order to define exterior powers within them? This structure should be there in the case of usual module categories. PS: I am interested in exterior powers since they should reduce locally free objects to line objects via the Plücker embedding. The$d$th anti-symmetric power of$1^n$is not$1^{\binom{n}{d}} = \wedge^d 1^n$in general, therefore it is not the correct notion here. EDIT: Here is a more specific formulation: Is there an$R[S_n]$-module$T$, such that for every$R$-module$X$, we have that$T \otimes_{R[S_n]} X^{\otimes n}$equals$\wedge^n X = X^{\otimes n}/(... \otimes x ... \otimes x ...)$? Probably not, because$T$should be a quotient of$R[S_n]$, and the corresponding ideal cannot generate squares. EDIT: Concerning the "hidden extra structure" in the case of modules, I have made the following observation: Let the ground ring be$\mathbb{Z}$(or more generally a ring$R$in which$r^2 - r \in 2R$for all$r \in R$; this indluces boolean rings such as$\mathbb{F}_2$and also$\mathbb{Z}/n$). Then there is a well-defined(!) map$f_n : X^{\otimes n} \to \text{Asym}(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$in the case of modules. The cokernel of$\oplus_{n \geq 2} f_n$is$\wedge(X)$. EDIT: If there won't be any definition in the general case, one might define$\lambda$tensor categories as categorified$\lambda$-rings. But I'm still curious why have quite often a distinguished$\lambda$structure, coming from a specific quotient of the tensor power ... 5 added 646 characters in body; edited body Let$C$be a cocomplete$R$-linear tensor category. Many notions in commutative algebra may be internalized to$C$. For example an algebra is an object$A$in$C$together with morphisms$e : 1 \to A$(unit) and$m: A \otimes A \to A$(multiplication) satisfying the usual laws. The$n$th symmetric power$\text{Sym}^n(X)$of an object$X$is the quotient of$X^{\otimes n}$by$x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of two maps$(X^{\otimes n})^{n!} \to X^{\otimes n}$. Then$\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$is an algebra object and$\text{Sym}$is in fact left adjoint to the forgetful functor from algebras in$C$to$C$. But now what about the exterior power$\wedge^n(X)$? It is clear how to define$X^{\otimes n}$modulo$x_1 \otimes ... \otimes x_n = \text{sign}(\sigma) x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$in this context, which one might call the skew-symmetric power. The correct definition of the exterior power also has to mod out$... \otimes a \otimes ... \otimes a \otimes ... = 0$. But I have no idea how to internalize this to$C$. The reason is that there is no morphism$X \to X \otimes X$which acts like$a \mapsto a \otimes a$. Another idea would be to define$\wedge(X)$as a graded-commutative algebra object with the usual universal property, classifying maps from$X$which satisfy something like$f(x)^2=0$, but again it is unclear how to formulate this in$C$. If this is not possible at all, which additional structure on$C$do we need in order to define exterior powers within them? This structure should be there in the case of usual module categories. PS: I am interested in exterior powers since they should reduce locally free objects to line objects via the Plücker embedding. The$d$th anti-symmetric power of$1^n$is not$1^{\binom{n}{d}} = \wedge^d 1^n$in general, therefore it is not the correct notion here. EDIT: Here is a more specific formulation: Is there an$R[S_n]$-module$T$, such that for every$R$-module$X$, we have that$T \otimes_{R[S_n]} X^{\otimes n}$equals$\wedge^n X = X^{\otimes n}/(... \otimes x ... \otimes x ...)$? Probably not, because$T$should be a quotient of$R[S_n]$, and the corresponding ideal cannot generate squares. EDIT: Concerning the "hidden extra structure" in the case of modules, I have made the following observation: Let the ground ring be$\mathbb{Z}$(or more generally a ring$R$in which$r^2 - r \in 2R$for all$r \in R$; this indluces boolean rings such as$\mathbb{F}_2$and also$\mathbb{Z}/n$). Then there is a well-defined(!) map$f_n : X^{\otimes n} \to \text{Asym}(X), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$in the case of modules. The cokernel of$\oplus_{n \geq 2} f_n$is$\wedge(X)\$.

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