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I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x\cdot x$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of Cl is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x\cdot x$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

Alberto García-Raboso gives an answer as well, where in the discussion it is settled that $Cl$ preserves finite coproducts. If we can also show that it preserves cokernels then we know that it must have a right-adjoint, right, by Freyd adjoint functor theorem?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form as a quotient of the tensor algebra. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x^2$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of $Cl$ is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x^2$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

Alberto García-Raboso gives an answer as well, where in the discussion it is settled that $Cl$ preserves finite coproducts. If we can also show that it preserves cokernels then we know that it must have a right-adjoint, by Freyd adjoint functor theorem, right?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x\cdot x$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of Cl is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x\cdot x$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x\cdot x$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of Cl is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x\cdot x$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

Alberto García-Raboso gives an answer as well, where in the discussion it is settled that $Cl$ preserves finite coproducts. If we can also show that it preserves cokernels then we know that it must have a right-adjoint, right, by Freyd adjoint functor theorem?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, it seems a plausible answer. But after some thought, I'm not convinced.

We know that a universal construction, if exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x\cdot x$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of Cl is a forgetful functor $k-\text{Alg}\to k-\text{Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x\cdot x$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.

I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.

We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.

An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x\cdot x$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.

I should conclude that the right-adjoint of Cl is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.

This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x\cdot x$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?

And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?

And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?

I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.